Applied lesson: heat phenomena in everyday life

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 16-1Core

   A sauna stove contains $20\,\text{kg}$ of stones with a specific heat capacity of $800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$. The stones warm from $20\,{}^\circ\text{C}$ to $300\,{}^\circ\text{C}$.

   How large a quantity of heat is transferred to the stones? Give your answer in megajoules.

   MJCheck
   Answer

   $4.5\,\text{MJ}$

   Show the solution

   Given

   stones $m = 20\,\text{kg}$, $T_\text{initial} = 20\,{}^\circ\text{C}$, $T_\text{final} = 300\,{}^\circ\text{C}$; given in the problem $c = 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$

   Find

   quantity of heat $Q$ transferred to the stones, in megajoules

   Principle

   The stones warm up but stay solid, so their state does not change and no latent heat is needed. Then the quantity of heat absorbed is proportional to the mass and to the change in temperature, and the constant of proportionality is the specific heat capacity $c$.

   Formula

   $Q = c\,m\,\Delta T$, where $\Delta T = T_\text{final} - T_\text{initial}$

   Conversions

   The change in temperature is the same in degrees Celsius and in kelvins:

   $\Delta T$$= 300\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 280\,\text{K}$

   Finally $1\,\text{MJ} = 10^6\,\text{J}$.

   Substitution

   $Q$$= 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 20\,\text{kg} \cdot 280\,\text{K}$

   $= 4\,480\,000\,\text{J}$

   $= 4.48\,\text{MJ}$

   Answer

   $4.5\,\text{MJ}$

   Two significant figures, because the mass $20\,\text{kg}$ is known to two figures.

   Reasonableness

   Estimate: $800 \cdot 20 \cdot 300\,\text{J} \approx 4.8\,\text{MJ}$, so the result is of the right order of magnitude. $Q$ is positive because the stones warm up. Heating the same mass of water by the same amount would take about five times the energy, because the $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ of water is about five times larger.

2. Exercise 16-2Core

   $0.20\,\text{kg}$ of water that is already at the boiling point of $100\,{}^\circ\text{C}$ is thrown on the sauna stones. The water vaporises completely.

   How large a quantity of heat is used for vaporisation? Give your answer in kilojoules.

   kJCheck
   Answer

   $450\,\text{kJ}$

   Show the solution

   Given

   water $m = 0.20\,\text{kg}$, already at the boiling point $100\,{}^\circ\text{C}$; from the table $r = 2260\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   quantity of heat $Q$ needed for vaporisation, in kilojoules

   Principle

   The water is already at $100\,{}^\circ\text{C}$ and is vaporising, so its temperature does not change: all the energy goes into the change of state and none into raising the temperature. So we use the specific latent heat of vaporisation $r$, not $c\,\Delta T$.

   Formula

   $Q = r\,m$

   Conversions

   Not needed: $r$ is in kilojoules per kilogram and the mass is in kilograms, so the product comes directly in kilojoules.

   Substitution

   $Q$$= 2260\,\tfrac{\text{kJ}}{\text{kg}} \cdot 0.20\,\text{kg}$

   $= 452\,\text{kJ}$

   Answer

   $450\,\text{kJ}$

   Two significant figures like the mass $0.20\,\text{kg}$.

   Reasonableness

   Vaporising a kilogram takes over two megajoules, so a fifth of a kilogram takes about $0.2 \cdot 2300\,\text{kJ} \approx 460\,\text{kJ}$. Heating the same amount of water from $20\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$ would take only $4190 \cdot 0.20 \cdot 80\,\text{J} \approx 67\,\text{kJ}$, so vaporisation takes about seven times as much.

3. Exercise 16-3

   The total area of the walls of a cool bag is $0.60\,\text{m}^2$ and the U-value is $1.6\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$. It is $2.0\,{}^\circ\text{C}$ inside the bag and $30\,{}^\circ\text{C}$ outside.

   How large is the heat power conducted through the walls into the bag? Give your answer in watts.

   WCheck
   Answer

   $27\,\text{W}$

   Show the solution

   Given

   $U = 1.6\,\tfrac{\text{W}}{\text{m}^2\text{K}}$, $A = 0.60\,\text{m}^2$, $T_\text{in} = 2.0\,{}^\circ\text{C}$, $T_\text{out} = 30\,{}^\circ\text{C}$

   Find

   power $P$ conducted through the walls into the bag, in watts

   Principle

   Heat is conducted through the wall from warm to cold, and when the temperature difference stays the same, the power stays constant. The power is proportional to the area of the walls and to the temperature difference, and the constant of proportionality is the $U$-value.

   Formula

   $P = U A\,\Delta T$, where $\Delta T = T_\text{out} - T_\text{in}$

   Conversions

   The temperature difference is the same in kelvins:

   $\Delta T$$= 30\,{}^\circ\text{C} - 2.0\,{}^\circ\text{C}$

   $= 28\,\text{K}$

   Nothing else is needed, because

   $\tfrac{\text{W}}{\text{m}^2\text{K}} \cdot \text{m}^2 \cdot \text{K}$$= \text{W}$

   Substitution

   $P$$= 1.6\,\tfrac{\text{W}}{\text{m}^2\text{K}} \cdot 0.60\,\text{m}^2 \cdot 28\,\text{K}$

   $= 26.88\,\text{W}$

   Answer

   $27\,\text{W}$

   Two significant figures like the values $1.6$ and $0.60$ and the difference $28\,\text{K}$.

   Reasonableness

   Estimate: $1.6 \cdot 0.6 \approx 1\,\tfrac{\text{W}}{\text{K}}$, and $28\,\text{K}$ gives about $28\,\text{W}$. The power is positive because heat comes in from outside. It is small because the wall insulates well, but in an hour the energy coming in is still $27\,\text{W} \cdot 3600\,\text{s} \approx 97\,\text{kJ}$, which would be enough to melt about $0.3\,\text{kg}$ of ice per hour.

4. Exercise 16-4

   A $0.75\,\text{kg}$ ice pack in a cool bag is at $0\,{}^\circ\text{C}$ and melts completely.

   How large a quantity of heat is transferred to the ice? Give your answer in kilojoules.

   kJCheck
   Answer

   $250\,\text{kJ}$

   Show the solution

   Given

   ice $m = 0.75\,\text{kg}$, at the melting point $0\,{}^\circ\text{C}$; from the table $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   quantity of heat $Q$ needed to melt the ice, in kilojoules

   Principle

   The ice is already at the melting point, and while it melts it stays at $0\,{}^\circ\text{C}$. So the energy does not raise the temperature but changes the interactions between the particles: the regular arrangement of the ice breaks down and the interaction energy between the particles increases, and we need the specific latent heat of fusion $s$, not $c\,\Delta T$.

   Formula

   $Q = s\,m$

   Conversions

   Not needed: $s$ is in kilojoules per kilogram and the mass is in kilograms.

   Substitution

   $Q$$= 334\,\tfrac{\text{kJ}}{\text{kg}} \cdot 0.75\,\text{kg}$

   $= 250.5\,\text{kJ}$

   Answer

   $250\,\text{kJ}$

   Two significant figures like the mass $0.75\,\text{kg}$.

   Reasonableness

   Estimate: a third of a megajoule per kilogram, so three quarters of a kilogram takes about $250\,\text{kJ}$. Melting takes the same energy as heating the same amount of water by about 80 kelvins: $4190 \cdot 0.75 \cdot 80\,\text{J} \approx 251\,\text{kJ}$.

5. Exercise 16-5

   A car's coolant ($c = 3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$) circulates in the engine at $0.50\,\text{kg}$ every second, and the liquid warms by $8.0\,\text{K}$ in the engine.

   How large is the power transferred from the engine to the liquid? Give your answer in kilowatts.

   kWCheck
   Answer

   $14\,\text{kW}$

   Show the solution

   Given

   $c = 3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, circulating mass $\tfrac{m}{t} = 0.50\,\tfrac{\text{kg}}{\text{s}}$, rise in temperature $\Delta T = 8.0\,\text{K}$

   Find

   power $P$ transferred from the engine to the liquid, in kilowatts

   Principle

   The liquid warms up and does not boil, so its state does not change, and the quantity of heat absorbed per second is $c\,m\,\Delta T$ for the mass that passes through the engine in a second. Power is energy per unit time, $P = \tfrac{Q}{t}$.

   Formula

   $P$$= \dfrac{Q}{t}$

   $= \dfrac{c\,m\,\Delta T}{t}$

   $= c\,\dfrac{m}{t}\,\Delta T$

   Conversions

   The change in temperature is already in kelvins. Units:

   $\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot \tfrac{\text{kg}}{\text{s}} \cdot \text{K}$$= \tfrac{\text{J}}{\text{s}}$

   $= \text{W}$

   and finally $1\,\text{kW} = 1000\,\text{W}$.

   Substitution

   $P$$= 3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 0.50\,\tfrac{\text{kg}}{\text{s}} \cdot 8.0\,\text{K}$

   $= 14\,000\,\tfrac{\text{J}}{\text{s}}$

   $= 14\,000\,\text{W}$

   $= 14\,\text{kW}$

   Answer

   $14\,\text{kW}$

   Two significant figures like the values $0.50\,\tfrac{\text{kg}}{\text{s}}$ and $8.0\,\text{K}$.

   Reasonableness

   Every second, $14\,\text{kJ}$ is transferred to the liquid. Estimate: $3500 \cdot 0.5 \cdot 8 = 14\,000$. The power carried by the coolant of a passenger car is typically tens of kilowatts, so the value is of the right order of magnitude.

6. Exercise 16-6

   A wooden door handle and a metal door handle have been in a room for a long time, and both are at room temperature. The metal handle still feels colder than the wooden one.

   Which of the following explains this?

   Choose an option.

   • The metal is at a lower temperature than the wood in the room, because it cannot keep heat inside itself.
   • The metal produces cold, which flows from the handle into the hand when the handle is touched for a moment.
   • The internal energy of the metal is smaller than that of the wood, so the metal is colder than the wood even in the room.
   • The metal conducts heat away from the skin faster than the wood does, so the temperature of the skin drops more.

   Answer
   I have done my solution.
   Answer

   The metal conducts heat away from the skin faster than the wood does, so the temperature of the skin drops more.

   Show the solution

   Find

   why a metal handle feels colder than a wooden handle even though both are at room temperature.

   Principle

   Objects that have been in a room for a long time are in thermal equilibrium with their surroundings, so the handles are at the same temperature. What decides the feel is how fast energy moves from the warm hand into the handle: because of the temperature difference there is a net transfer of energy from the hand to the handle, and the rate of the transfer depends on the thermal conductivity of the material.

   Model answer

   Metal conducts heat well, so a lot of energy moves quickly from the skin of the hand into the metal and the temperature of the skin drops quickly. Wood conducts heat poorly, so energy moves more slowly and the surface of the skin stays warmer. So what we sense is the temperature of the skin and the rate at which energy is transferred, not the temperature of the handle. Cold does not flow from the handle into the hand, and the internal energy of the handle does not decide the matter: both are at the temperature of the room.

7. Exercise 16-7

   Heat is conducted through the wall of a cool bag from outside into the ice pack all the time. Still, the temperature inside the bag stays at $0\,{}^\circ\text{C}$ until the ice has melted.

   Which of the following explains this?

   Choose an option.

   • The walls of the bag prevent heat from being conducted at all, so no energy enters the bag before the ice melts.
   • The specific heat capacity of ice is so large that the energy entering the bag does not raise the temperature at all.
   • The energy entering the bag melts ice and does not raise the temperature until all the ice has melted.
   • The ice produces cold, which cancels the heat entering the bag and keeps the temperature unchanged.

   Answer
   I have done my solution.
   Answer

   The energy entering the bag melts ice and does not raise the temperature until all the ice has melted.

   Show the solution

   Find

   why the temperature inside the bag stays at $0\,{}^\circ\text{C}$ although energy comes through the wall all the time.

   Principle

   A power $P = UA\Delta T$ is transferred into the bag through the wall, and its energy $P\,t$ goes to the ice. The ice is at its melting point, so the energy does not raise the temperature but is used for melting, $Q = s\,m$: it increases the interaction energy between the particles. The temperature can rise only after all the ice has melted.

   Model answer

   The energy entering the bag melts ice, and the temperature does not change during melting. So it is $0\,{}^\circ\text{C}$ inside as long as there is ice left. Ice does not produce cold; it cools by taking in energy. The wall also only slows down the conduction of heat and does not prevent it, and it is exactly the energy coming through the wall that melts the ice. When the ice has melted, the incoming energy warms the water and the temperature starts to rise.

8. Exercise 16-8

   An electric kettle has a power of $2.0\,\text{kW}$ and contains $1.5\,\text{kg}$ of water at $20\,{}^\circ\text{C}$ with a specific heat capacity of $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$. Assume that all of the electric power goes into heating the water and that the water does not yet boil.

   How long does it take for the water to warm to $100\,{}^\circ\text{C}$? Give your answer in seconds.

   sCheck
   Answer

   $250\,\text{s}$

   Show the solution

   Given

   power $P = 2.0\,\text{kW}$, water $m = 1.5\,\text{kg}$, $T_\text{initial} = 20\,{}^\circ\text{C}$, $T_\text{final} = 100\,{}^\circ\text{C}$; given in the problem $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$

   Find

   heating time $t$ in seconds

   Principle

   By the assumption, all of the electrical energy $P\,t$ goes into heating the water. The water does not boil or vaporise, so its state does not change and $Q = c\,m\,\Delta T$ is enough. From the energy balance $P\,t = c\,m\,\Delta T$.

   Formula

   $P\,t = c\,m\,\Delta T$, so $t = \dfrac{c\,m\,\Delta T}{P}$

   Conversions

   The change in temperature is the same in kelvins:

   $\Delta T$$= 100\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   The power is $P = 2.0\,\text{kW} = 2000\,\text{W}$, so that the energy is in joules and the time in seconds.

   Substitution

   $t$$= \dfrac{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 1.5\,\text{kg} \cdot 80\,\text{K}}{2000\,\text{W}}$

   $= \dfrac{502\,800\,\text{J}}{2000\,\text{W}}$

   $= 251.4\,\text{s}$

   Answer

   $250\,\text{s}$

   Two significant figures, as in the mass $1.5\,\text{kg}$ and the power $2.0\,\text{kW}$.

   Reasonableness

   Estimate: $4000 \cdot 1.5 \cdot 80\,\text{J} \approx 480\,\text{kJ}$, and $480\,\text{kJ} : 2\,\text{kW} = 240\,\text{s}$, about four minutes, which is the right order of magnitude for an ordinary kettle. Check with the energy balance: $2000\,\text{W} \cdot 251.4\,\text{s} \approx 503\,\text{kJ}$ is the same as the $503\,\text{kJ}$ received by the water. The real time would be slightly longer, because part of the energy warms the kettle and the air.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 16-9Core

   The total area of the walls of a cool bag is $0.50\,\text{m}^2$ and the U-value is $2.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$. It is $25\,{}^\circ\text{C}$ outside, and the temperature of the ice pack stays at $0\,{}^\circ\text{C}$ as long as there is ice.

   How large must the mass of the ice pack be for the ice to last for 6.0 hours? Assume that the bag contains only ice.

   kgCheck
   Answer

   $1.6\,\text{kg}$

   Show the solution

   Given

   $U = 2.0\,\tfrac{\text{W}}{\text{m}^2\text{K}}$, $A = 0.50\,\text{m}^2$, outside $25\,{}^\circ\text{C}$ and inside $0\,{}^\circ\text{C}$, so $\Delta T = 25\,\text{K}$, time $t = 6.0\,\text{h}$; from the table $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   mass of the ice $m$ in kilograms

   Principle

   As long as there is ice, the inside stays at $0\,{}^\circ\text{C}$, so the temperature difference and the power through the wall stay constant. The bag contains only ice, so all the energy $P\,t$ that has come through the wall is used to melt the ice: $P\,t = s\,m$.

   Formula

   $U A\,\Delta T\,t = s\,m$, so $m = \dfrac{U A\,\Delta T\,t}{s}$

   Conversions

   $t$$= 6.0\,\text{h}$

   $= 6.0 \cdot 3600\,\text{s}$

   $= 21\,600\,\text{s}$

   $s$$= 334\,\tfrac{\text{kJ}}{\text{kg}}$

   $= 334\,000\,\tfrac{\text{J}}{\text{kg}}$

   so that a watt times a second is a joule.

   Substitution

   $P$$= 2.0 \cdot 0.50 \cdot 25\,\text{W}$

   $= 25\,\text{W}$

   $m$$= \dfrac{25\,\text{W} \cdot 21\,600\,\text{s}}{334\,000\,\tfrac{\text{J}}{\text{kg}}}$

   $= \dfrac{540\,000\,\text{J}}{334\,000\,\tfrac{\text{J}}{\text{kg}}}$

   $= 1.617\,\text{kg}$

   Answer

   $1.6\,\text{kg}$

   Two significant figures, because the U-value $2.0$, the area $0.50\,\text{m}^2$, the temperature difference $25\,\text{K}$ and the time $6.0\,\text{h}$ all have two figures.

   Reasonableness

   The power is $25\,\text{W}$, so in six hours $540\,\text{kJ}$ comes into the bag. Check: $1.617\,\text{kg} \cdot 334\,\tfrac{\text{kJ}}{\text{kg}} = 540\,\text{kJ}$. An ice pack of a little over one and a half kilograms is a common size, and twice the time would need twice the mass.

2. Exercise 16-10Core

   $0.25\,\text{kg}$ of water at $20\,{}^\circ\text{C}$ is thrown on the stones as sauna water. The water warms to $100\,{}^\circ\text{C}$ and vaporises completely.

   How large is the total quantity of heat used? Give your answer in kilojoules.

   kJCheck
   Answer

   $650\,\text{kJ}$

   Show the solution

   Given

   water $m = 0.25\,\text{kg}$, $T_\text{initial} = 20\,{}^\circ\text{C}$, $T_\text{final} = 100\,{}^\circ\text{C}$; from the table $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and $r = 2260\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   total quantity of heat $Q$ in kilojoules

   Principle

   The process has two stages, one after the other. First the water warms up to $100\,{}^\circ\text{C}$ ($c\,m\,\Delta T$), then it vaporises at this temperature, which does not change ($r\,m$). There is no change in temperature during vaporisation, so the energies of the stages are calculated with different formulas and added together.

   Formula

   $Q$$= Q_\text{heating} + Q_\text{vaporisation}$

   $= c\,m\,\Delta T + r\,m$

   where $\Delta T = T_\text{final} - T_\text{initial}$

   Conversions

   $\Delta T$$= 100\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   the same in kelvins.

   $r$$= 2260\,\tfrac{\text{kJ}}{\text{kg}}$

   $= 2\,260\,000\,\tfrac{\text{J}}{\text{kg}}$

   and finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $Q_\text{heating}$

   $= 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 0.25\,\text{kg} \cdot 80\,\text{K}$

   $= 83\,800\,\text{J}$

   $Q_\text{vaporisation}$

   $= 2\,260\,000\,\tfrac{\text{J}}{\text{kg}} \cdot 0.25\,\text{kg}$

   $= 565\,000\,\text{J}$

   so

   $Q$$= 648\,800\,\text{J}$

   $= 648.8\,\text{kJ}$

   Answer

   $650\,\text{kJ}$

   Two significant figures like the mass $0.25\,\text{kg}$.

   Reasonableness

   Vaporisation takes $\tfrac{565}{648.8} \approx 87\,\%$ of the total energy, even though the temperature does not change during it. Estimate: $0.25\,\text{kg}$ needs about $2.6\,\tfrac{\text{MJ}}{\text{kg}}$ per kilogram, so about $650\,\text{kJ}$. If vaporisation were forgotten, the result would be only $84\,\text{kJ}$.

3. Exercise 16-11

   An engine transfers a power of $24\,\text{kW}$ to the coolant ($c = 3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$). The liquid may warm by at most $8.0\,\text{K}$ in the engine.

   What mass of liquid has to be circulated through the engine every second? Give your answer in kg/s.

   kg/sCheck
   Answer

   $0.86\,\tfrac{\text{kg}}{\text{s}}$

   Show the solution

   Given

   $P = 24\,\text{kW}$, $c = 3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, greatest allowed rise in temperature $\Delta T = 8.0\,\text{K}$

   Find

   mass of liquid to be circulated per second $\tfrac{m}{t}$ in $\tfrac{\text{kg}}{\text{s}}$

   Principle

   All the power transferred from the engine goes into warming the liquid, and the liquid does not boil, so its state does not change. The mass $\tfrac{m}{t}$ passing per second absorbs the quantity of heat $c\,\tfrac{m}{t}\,\Delta T$, and this is equal to the power $P$.

   Formula

   $P = c\,\dfrac{m}{t}\,\Delta T$, so $\dfrac{m}{t} = \dfrac{P}{c\,\Delta T}$

   Conversions

   $24\,\text{kW}$$= 24\,000\,\text{W}$

   $= 24\,000\,\tfrac{\text{J}}{\text{s}}$

   The change in temperature is already in kelvins.

   Substitution

   $\dfrac{m}{t}$$= \dfrac{24\,000\,\tfrac{\text{J}}{\text{s}}}{3500\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 8.0\,\text{K}}$

   $= \dfrac{24\,000}{28\,000}\,\tfrac{\text{kg}}{\text{s}}$

   $= 0.857\,\tfrac{\text{kg}}{\text{s}}$

   Answer

   $0.86\,\tfrac{\text{kg}}{\text{s}}$

   Two significant figures like the values $24\,\text{kW}$ and $8.0\,\text{K}$.

   Reasonableness

   Check: $3500 \cdot 0.857 \cdot 8.0\,\text{W} = 24\,\text{kW}$. With pure water ($c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$) about $0.72\,\tfrac{\text{kg}}{\text{s}}$ would be enough, so a liquid containing glycol, with a smaller $c$, needs a mass flow about a fifth larger. The direction is right: the worse the liquid stores energy, the more of it has to be circulated.

4. Exercise 16-12

   To make a cold drink, $0.33\,\text{kg}$ of lemonade at $22\,{}^\circ\text{C}$, with a specific heat capacity equal to that of water, is poured into a glass. Enough ice is added to the glass.

   How many grams of ice melt when the lemonade cools to $0\,{}^\circ\text{C}$? Assume that no heat is transferred from the surroundings.

   gCheck
   Answer

   $91\,\text{g}$

   Show the solution

   Given

   lemonade $m_\text{j} = 0.33\,\text{kg}$, $T_\text{initial} = 22\,{}^\circ\text{C}$, $T_\text{final} = 0\,{}^\circ\text{C}$, $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ (the value for water); from the table $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   mass of melted ice $m_\text{ice}$ in grams

   Principle

   Energy is conserved: when no heat comes in from the surroundings, the energy that the lemonade gives out as it cools is all used to melt ice. There is enough ice, so the temperature settles at $0\,{}^\circ\text{C}$ and the melted ice does not warm up.

   Formula

   $c\,m_\text{j}\,\Delta T = s\,m_\text{ice}$, so

   $m_\text{ice}$$= \dfrac{c\,m_\text{j}\,\Delta T}{s}$

   Conversions

   $\Delta T$$= 22\,{}^\circ\text{C} - 0\,{}^\circ\text{C}$

   $= 22\,\text{K}$

   the same in kelvins. $s = 334\,000\,\tfrac{\text{J}}{\text{kg}}$, and finally $1\,\text{kg} = 1000\,\text{g}$.

   Substitution

   $m_\text{ice}$$= \dfrac{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 0.33\,\text{kg} \cdot 22\,\text{K}}{334\,000\,\tfrac{\text{J}}{\text{kg}}}$

   $= \dfrac{30\,419.4\,\text{J}}{334\,000\,\tfrac{\text{J}}{\text{kg}}}$

   $= 0.0911\,\text{kg}$

   $= 91.1\,\text{g}$

   Answer

   $91\,\text{g}$

   Two significant figures like the values $0.33\,\text{kg}$ and $22\,\text{K}$.

   Reasonableness

   The mass of melted ice is $\tfrac{91}{330} \approx 28\,\%$ of the mass of the lemonade, because melting needs the energy of heating by about 80 kelvins per kilogram: $\tfrac{22}{80} \approx 0.28$. The mass is positive and smaller than that of the lemonade, as it should be.

5. Exercise 16-13

   The heating elements of a sauna stove are switched on at a power of $6.0\,\text{kW}$, and the stove contains $40\,\text{kg}$ of stones ($c = 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$). The table gives sample data on the temperature of the stones during heating. The efficiency of the heating is the energy transferred to the stones divided by the electrical energy used by the elements.

   a) Calculate the efficiency for the whole $30$ minutes of heating, as a percentage.

   b) During each $10$-minute period the temperature of the stones rises less than during the previous one. Explain why.

   Give the result of part a) in the answer box.

   Temperature of the sauna stones during heating (sample data)

   $t\,(\text{min})$	$T\,({}^\circ\text{C})$
   0	20
   10	112
   20	196
   30	275

   %Check
   Answer

   $76\,\%$

   Show the solution

   Given

   power $P = 6.0\,\text{kW}$, stones $m = 40\,\text{kg}$ and $c = 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$; from the table $T = 20\,{}^\circ\text{C}$ at $0\,\text{min}$ and $T = 275\,{}^\circ\text{C}$ at $30\,\text{min}$ (intermediate values $112\,{}^\circ\text{C}$ and $196\,{}^\circ\text{C}$)

   Find

   efficiency $\eta$ of the heating as a percentage

   Principle

   The stones stay solid, so the energy transferred to them is $Q = c\,m\,\Delta T$. The elements convert electrical energy into heat at the power $P$, so the energy used is $P\,t$. Efficiency is the ratio of the useful energy to the energy used.

   Formula

   $\eta$$= \dfrac{Q}{P\,t}$

   $= \dfrac{c\,m\,\Delta T}{P\,t}$

   Conversions

   The change in temperature is the same in kelvins:

   $\Delta T$$= 275\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 255\,\text{K}$

   The time is $t = 30\,\text{min} = 1800\,\text{s}$ and the power is $P = 6.0\,\text{kW} = 6000\,\text{W}$.

   Substitution

   $\eta$$= \dfrac{800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 40\,\text{kg} \cdot 255\,\text{K}}{6000\,\text{W} \cdot 1800\,\text{s}}$

   $= \dfrac{8.16 \cdot 10^{6}\,\text{J}}{10.8 \cdot 10^{6}\,\text{J}}$

   $= 0.756$

   $= 75.6\,\%$

   Answer

   $76\,\%$

   Two significant figures, as in the power $6.0\,\text{kW}$ and the mass $40\,\text{kg}$.

   Reasonableness

   The efficiency is below $100\,\%$, as it must be. If all the energy ended up in the stones, the temperature would rise by

   $\dfrac{10.8 \cdot 10^{6}\,\text{J}}{800 \cdot 40\,\text{J/K}}$$\approx 340\,\text{K}$

   whereas the measured rise is $255\,\text{K}$, about three quarters of that. Period by period, the energy $c\,m\,\Delta T$ is $2.94\,\text{MJ}$, $2.69\,\text{MJ}$ and $2.53\,\text{MJ}$, while $3.6\,\text{MJ}$ of electrical energy is used per period, so the efficiencies of the periods are about $82\,\%$, $75\,\%$ and $70\,\%$.

   Model answer for part b)

   the stones give out heat to the surroundings, and the power lost grows as the temperature difference grows. As the stones get hotter, a larger share of the elements' power of $6.0\,\text{kW}$ escapes to the surroundings and a smaller share warms the stones. So the rise slows down period by period ($92\,\text{K}$, $84\,\text{K}$, $79\,\text{K}$).

6. Exercise 16-14

   The table gives sample data on the heating energy $E$ used per month by a detached house and the mean outdoor temperature of the month, $T_\text{out}$. Assume that the heating energy equals the energy lost through the structures, that the indoor temperature is the same every month, and that a month has $720\,\text{h}$.

   a) Draw a graph of the points $(T_\text{out}, E)$ and fit a straight line to it.

   b) From the slope of the line, determine the heat loss coefficient $UA$ of the house in W/K, when the power is $P = UA\Delta T$.

   c) At what outdoor temperature does the line cross the horizontal axis, and what does it tell you about the house?

   Give the result of part b) in the answer box.

   Heating energy per month (sample data)

   $T_\text{out}\,({}^\circ\text{C})$	$E\,(\text{kWh})$
   8	1210
   2	1790
   −3	2340
   −8	2790

   W/KCheck
   Answer

   $1.4 \cdot 10^{2}\,\dfrac{\text{W}}{\text{K}}$

   Show the solution

   Given

   from the table, points $(T_\text{out}$; $E)$: $(8\,{}^\circ\text{C}$; $1210\,\text{kWh})$, $(2\,{}^\circ\text{C}$; $1790\,\text{kWh})$, $(-3\,{}^\circ\text{C}$; $2340\,\text{kWh})$, $(-8\,{}^\circ\text{C}$; $2790\,\text{kWh})$; time in a month $t = 720\,\text{h}$

   Find

   heat loss coefficient $UA$ of the house in $\text{W/K}$

   Principle

   The heating energy equals the energy lost through the structures, $E = P\,t = UA\,\Delta T\,t$, and $\Delta T = T_\text{in} - T_\text{out}$. When the indoor temperature and $UA$ are constants, $E$ is a linear function of the outdoor temperature, and the slope of the line is $-UA\,t$.

   Reading the graph

   The points $(T_\text{out}, E)$ lie approximately on a descending line, but not exactly on it because of variation in the measurements. The fitted line passes between the points. The slope is read from two points of the line: $(8\,{}^\circ\text{C}$; $1210\,\text{kWh})$ and $(-8\,{}^\circ\text{C}$; $2810\,\text{kWh})$ (a point of the line, not a measured point; the value in the table is $2790\,\text{kWh}$), so

   $k$$= \dfrac{2810\,\text{kWh} - 1210\,\text{kWh}}{-8\,\text{K} - 8\,\text{K}}$

   $= -100\,\dfrac{\text{kWh}}{\text{K}}$

   Formula

   $k = -UA\,t$, so $UA = \dfrac{-k}{t}$

   Conversions

   The magnitude of the slope is $100\,\dfrac{\text{kWh}}{\text{K}}$, and

   $\dfrac{\text{kWh}}{\text{h}}$$= \text{kW}$

   Finally $1\,\text{kW} = 1000\,\text{W}$.

   Substitution

   $UA$$= \dfrac{100\,\tfrac{\text{kWh}}{\text{K}}}{720\,\text{h}}$

   $= 0.1389\,\dfrac{\text{kW}}{\text{K}}$

   $= 139\,\dfrac{\text{W}}{\text{K}}$

   Answer

   $1.4 \cdot 10^{2}\,\dfrac{\text{W}}{\text{K}}$

   (calculated value $139\,\dfrac{\text{W}}{\text{K}}$), two significant figures, because fitting the line limits the accuracy.

   Reasonableness

   Check against the data: in a month like February, when $T_\text{out} = -8\,{}^\circ\text{C}$ and $\Delta T \approx 28\,\text{K}$, the power is $139 \cdot 28\,\text{W} \approx 3.9\,\text{kW}$ and the energy is $3.9\,\text{kW} \cdot 720\,\text{h} \approx 2800\,\text{kWh}$, which matches the value $2790\,\text{kWh}$ in the table.

   Model answer for part c)

   the line crosses the horizontal axis when $E = 0$, that is, when $\Delta T = 0$: at about $+20\,{}^\circ\text{C}$, which is the indoor temperature of the house. Above it no heating is needed, so the line tells you the indoor temperature, not a random number. In a real house, people, appliances and the sun produce free heat, so the intersection is typically lower than the indoor temperature.

7. Exercise 16-15

   A student calculates how long $0.80\,\text{kg}$ of ice lasts in a cool bag. The total area of the walls of the bag is $0.70\,\text{m}^2$ and the U-value is $1.4\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, the ice is at $0\,{}^\circ\text{C}$, and it is $26\,{}^\circ\text{C}$ outside. The bag contains only ice, and the latent heat of fusion of ice is $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$. The student's solution is this:

   “Temperatures must be converted to kelvins, so $\Delta T = 26 + 273 = 299\,\text{K}$. The power is $P = UA\Delta T = 0.98 \cdot 299\,\text{W} \approx 293\,\text{W}$ and the time is $t = \dfrac{s\,m}{P} \approx 910\,\text{s} \approx 0.25\,\text{h}$.”

   a) What is wrong in the student's solution?

   b) Calculate correctly how long the ice lasts, in hours.

   Give the result of part b) in the answer box.

   hCheck
   Answer

   $2.9\,\text{h}$

   Show the solution

   Given

   ice $m = 0.80\,\text{kg}$, $T_\text{ice} = 0\,{}^\circ\text{C}$, outside $26\,{}^\circ\text{C}$, $A = 0.70\,\text{m}^2$, $U = 1.4\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$; given in the problem $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   the time $t$ the ice lasts, in hours; also where the student's error is

   Principle

   The student used the outdoor temperature in kelvins as the temperature difference, $26 + 273 = 299\,\text{K}$. The formula $P = UA\Delta T$ needs the difference of the temperatures, and the difference is the same in degrees Celsius and in kelvins, because the scales have the same degree: $26\,{}^\circ\text{C} - 0\,{}^\circ\text{C} = 26\,\text{K}$, or $299\,\text{K} - 273\,\text{K}$. The student ignored the indoor temperature. When there is ice, it is $0\,{}^\circ\text{C}$ inside and the energy coming through the wall melts the ice, $UA\Delta T\,t = s\,m$.

   Formula

   $UA\Delta T\,t = s\,m$, so $t = \dfrac{s\,m}{UA\Delta T}$

   Conversions

   No conversion of temperatures: $\Delta T = 26\,\text{K}$. The latent heat of fusion is

   $s$$= 334\,\tfrac{\text{kJ}}{\text{kg}}$

   $= 334\,000\,\tfrac{\text{J}}{\text{kg}}$

   and finally $1\,\text{h} = 3600\,\text{s}$.

   Substitution

   $t$$= \dfrac{334\,000\,\tfrac{\text{J}}{\text{kg}} \cdot 0.80\,\text{kg}}{1.4\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 0.70\,\text{m}^2 \cdot 26\,\text{K}}$

   $= \dfrac{267\,200\,\text{J}}{25.48\,\text{W}}$

   $= 10\,487\,\text{s}$

   $= 2.913\,\text{h}$

   Answer

   $2.9\,\text{h}$

   Two significant figures, as in the values $0.80\,\text{kg}$, $0.70\,\text{m}^2$ and $1.4\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$.

   Reasonableness

   The student's result of $0.25\,\text{h}$, about $15$ minutes, would be too short: a power of $25\,\text{W}$ melts a kilogram of ice only in hours. The ratio of the wrong and correct results, $\dfrac{2.913}{0.253} \approx 11.5$, is the same as the ratio of the temperature differences, $\dfrac{299}{26} \approx 11.5$, which shows that the whole difference comes from the wrong $\Delta T$. The result is of the same order of magnitude as in the example of the section: about three hours.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 16-16

   A sauna stove contains $25\,\text{kg}$ of stones ($c = 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$). $0.20\,\text{kg}$ of water at 20 °C is thrown on the stones, and it warms to $100\,{}^\circ\text{C}$ and vaporises completely. Assume that all the energy needed is taken from the stones and that the warming of the steam is not taken into account.

   By how many kelvins does the temperature of the stones drop?

   KCheck
   Answer

   $26\,\text{K}$

   Show the solution

   Given

   stones $m_\text{k} = 25\,\text{kg}$, $c_\text{k} = 800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$; water $m_\text{v} = 0.20\,\text{kg}$, $T_\text{initial} = 20\,{}^\circ\text{C}$, warms to $100\,{}^\circ\text{C}$ and vaporises; from the table $c_\text{v} = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and $r = 2260\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   drop in the temperature of the stones $\Delta T_\text{k}$ in kelvins

   Principle

   Energy is conserved: when all the energy is taken from the stones, the energy given out by the stones is equal to the energy absorbed by the water. The water absorbs energy in two stages (warming up and vaporisation), and the stones only cool down, so their state does not change. The warming of the steam and the heating of the stove are not taken into account.

   Formula

   $c_\text{k}\,m_\text{k}\,\Delta T_\text{k}$$= m_\text{v}(c_\text{v}\,\Delta T_\text{v} + r)$

   so

   $\Delta T_\text{k}$$= \dfrac{m_\text{v}(c_\text{v}\,\Delta T_\text{v} + r)}{c_\text{k}\,m_\text{k}}$

   Conversions

   $\Delta T_\text{v}$$= 100\,{}^\circ\text{C} - 20\,{}^\circ\text{C}$

   $= 80\,\text{K}$

   the same in kelvins.

   $r$$= 2260\,\tfrac{\text{kJ}}{\text{kg}}$

   $= 2\,260\,000\,\tfrac{\text{J}}{\text{kg}}$

   Substitution

   $c_\text{v}\,\Delta T_\text{v}$$= 4190 \cdot 80\,\tfrac{\text{J}}{\text{kg}}$

   $= 335\,200\,\tfrac{\text{J}}{\text{kg}}$

   $Q_\text{v}$$= 0.20\,\text{kg} \cdot (335\,200 + 2\,260\,000)\,\tfrac{\text{J}}{\text{kg}}$

   $= 519\,040\,\text{J}$

   $\Delta T_\text{k}$$= \dfrac{519\,040\,\text{J}}{800\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 25\,\text{kg}}$

   $= 25.95\,\text{K}$

   Answer

   $26\,\text{K}$

   Two significant figures like the masses $0.20\,\text{kg}$ and $25\,\text{kg}$.

   Reasonableness

   Vaporisation takes $452\,\text{kJ}$, which is $87\,\%$ of the energy, and warming alone would lower the temperature of the stones by only about $3.4\,\text{K}$. The drop in the stones is small compared with their temperature of $300\,{}^\circ\text{C}$, because there are a lot of stones. Warming the steam would make the drop slightly larger, and a stove that keeps heating the stones would make it smaller; $26\,\text{K}$ is the drop without heating by the stove.

2. Exercise 16-17

   $0.500\,\text{kg}$ of ice at $0\,{}^\circ\text{C}$ and $2.00\,\text{kg}$ of water at $30.0\,{}^\circ\text{C}$ are mixed in an insulated container.

   What is the temperature of the mixture when thermal equilibrium has been reached? First check whether all the ice melts.

   °CCheck
   Answer

   $8.1\,{}^\circ\text{C}$

   Show the solution

   Given

   ice $m_\text{ice} = 0.500\,\text{kg}$, $0\,{}^\circ\text{C}$; water $m_\text{v} = 2.00\,\text{kg}$, $T_\text{v} = 30.0\,{}^\circ\text{C}$; the container is insulated; from the table $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   final temperature of the mixture $T$ in degrees Celsius

   Principle

   Energy is conserved, because the container is insulated: the energy given out by the water melts the ice and warms the melted water to the final temperature. This works only if all the ice melts. First check: as it cools to $0\,{}^\circ\text{C}$ the water can give out at most $c\,m_\text{v}\,T_\text{v} = 251\,400\,\text{J}$, and melting the ice needs only $s\,m_\text{ice} = 167\,000\,\text{J}$. Since $251\,400\,\text{J} > 167\,000\,\text{J}$, all the ice melts and the final temperature is above $0\,{}^\circ\text{C}$.

   Formula

   $c\,m_\text{v}(T_\text{v} - T) = s\,m_\text{ice} + c\,m_\text{ice}\,T$, so

   $T$$= \dfrac{c\,m_\text{v}\,T_\text{v} - s\,m_\text{ice}}{c\,(m_\text{v} + m_\text{ice})}$

   Conversions

   The ice is at $0\,{}^\circ\text{C}$, so the change in temperature of the melted water is $T - 0\,{}^\circ\text{C} = T$ and the water cools by $T_\text{v} - T$; the differences are the same in kelvins, and $T$ comes directly in degrees Celsius. $s = 334\,000\,\tfrac{\text{J}}{\text{kg}}$.

   Substitution

   The numerator is

   $4190 \cdot 2.00 \cdot 30.0\,\text{J} - 334\,000 \cdot 0.500\,\text{J}$

   $= 251\,400\,\text{J} - 167\,000\,\text{J}$

   $= 84\,400\,\text{J}$

   and the denominator is

   $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 2.50\,\text{kg}$

   $= 10\,475\,\tfrac{\text{J}}{\text{K}}$

   so

   $$T = \dfrac{84\,400\,\text{J}}{10\,475\,\tfrac{\text{J}}{\text{K}}} = 8.057\,{}^\circ\text{C}$$

   Answer

   $8.1\,{}^\circ\text{C}$

   Two significant figures. The initial values $0.500\,\text{kg}$, $2.00\,\text{kg}$ and $30.0\,{}^\circ\text{C}$ have three figures, but the subtraction in the numerator $251\,400 - 167\,000 = 84\,400\,\text{J}$ loses one figure, so two figures are left in the result.

   Reasonableness

   The result is between $0\,{}^\circ\text{C}$ and $30\,{}^\circ\text{C}$, and closer to the temperature of the ice, because melting also cools the water. Energy balance: the water gives out $4190 \cdot 2.00 \cdot (30.0 - 8.057)\,\text{J} = 183\,880\,\text{J}$, and the ice takes in

   $167\,000\,\text{J} + 4190 \cdot 0.500 \cdot 8.057\,\text{J}$

   $= 183\,880\,\text{J}$

   If the energy of the water had been less than $167\,000\,\text{J}$, the result would be $0\,{}^\circ\text{C}$ and some ice would be left.

3. Exercise 16-18

   The total area of the walls of a cool bag is $0.50\,\text{m}^2$ and the U-value is $1.5\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$. $0.50\,\text{kg}$ of drinks at $24\,{}^\circ\text{C}$, with a specific heat capacity equal to that of water, $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, is packed into the bag, along with an ice pack taken from the freezer at $-10\,{}^\circ\text{C}$ ($c_\text{ice} = 2100\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$).

   It is $30\,{}^\circ\text{C}$ outside.

   The ice must last for 6.0 hours.

   Assume that the drinks cool to $0\,{}^\circ\text{C}$ at once, that the ice first warms to $0\,{}^\circ\text{C}$ and then melts, and that the same power comes through the wall all the time as if the inside were at $0\,{}^\circ\text{C}$.

   What is the smallest mass the ice pack must have? Give your answer in kilograms.

   kgCheck
   Answer

   $1.5\,\text{kg}$

   Show the solution

   Given

   $U = 1.5\,\tfrac{\text{W}}{\text{m}^2\text{K}}$, $A = 0.50\,\text{m}^2$, outside $30\,{}^\circ\text{C}$ and inside $0\,{}^\circ\text{C}$, so $\Delta T = 30\,\text{K}$, $t = 6.0\,\text{h}$; drinks $m_\text{j} = 0.50\,\text{kg}$, $\Delta T_\text{j} = 24\,\text{K}$, $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$; ice $-10\,{}^\circ\text{C} \to 0\,{}^\circ\text{C}$, $c_\text{ice} = 2100\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$; from the table $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   smallest mass $m$ of the ice pack in kilograms

   Principle

   Energy is conserved: during the 6.0 hours the ice has to absorb the energy coming through the wall and the energy given out by the drinks. The ice first warms from $-10\,{}^\circ\text{C}$ to $0\,{}^\circ\text{C}$ ($c_\text{ice}\,\Delta T_\text{ice}$) and then melts ($s$). By the assumption of the problem the inside is at $0\,{}^\circ\text{C}$, so the power through the wall stays constant.

   Formula

   $m\,(c_\text{ice}\,\Delta T_\text{ice} + s)$

   $= U A\,\Delta T\,t + c\,m_\text{j}\,\Delta T_\text{j}$

   so

   $m$$= \dfrac{U A\,\Delta T\,t + c\,m_\text{j}\,\Delta T_\text{j}}{c_\text{ice}\,\Delta T_\text{ice} + s}$

   Conversions

   $t = 6.0\,\text{h} = 21\,600\,\text{s}$ and $s = 334\,000\,\tfrac{\text{J}}{\text{kg}}$. The change in the temperature of the ice is

   $\Delta T_\text{ice}$$= 0\,{}^\circ\text{C} - (-10\,{}^\circ\text{C})$

   $= 10\,\text{K}$

   the same in kelvins.

   Substitution

   Energy from the wall:

   $U A\,\Delta T\,t$$= 1.5 \cdot 0.50 \cdot 30 \cdot 21\,600\,\text{J}$

   $= 486\,000\,\text{J}$

   Energy from the drinks:

   $c\,m_\text{j}\,\Delta T_\text{j}$$= 4190 \cdot 0.50 \cdot 24\,\text{J}$

   $= 50\,280\,\text{J}$

   Energy taken by the ice per kilogram:

   $c_\text{ice}\,\Delta T_\text{ice} + s$

   $= 21\,000 + 334\,000$

   $= 355\,000\,\tfrac{\text{J}}{\text{kg}}$

   Smallest mass:

   $m$$= \dfrac{486\,000\,\text{J} + 50\,280\,\text{J}}{355\,000\,\tfrac{\text{J}}{\text{kg}}}$

   $= 1.511\,\text{kg}$

   Answer

   $1.5\,\text{kg}$

   Two significant figures, because the U-value $1.5$, the area $0.50\,\text{m}^2$, the mass of the drinks $0.50\,\text{kg}$ and the time $6.0\,\text{h}$ all have two figures.

   Reasonableness

   The wall brings in $486\,\text{kJ}$ and the drinks $50\,\text{kJ}$, about $536\,\text{kJ}$ in total. If the ice were already at $0\,{}^\circ\text{C}$, we would need $\tfrac{536\,\text{kJ}}{334\,\text{kJ/kg}} \approx 1.6\,\text{kg}$. Warming freezer ice takes about $\tfrac{21}{334} \approx 6\,\%$ more energy per kilogram, so the mass needed is slightly smaller, $1.51\,\text{kg}$, which with two figures gives the smallest suitable mass $1.5\,\text{kg}$.

4. Exercise 16-19

   The average U-value of the walls, roof and windows of a cottage is $0.50\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$ and their total area is $120\,\text{m}^2$. The indoor temperature is kept constant at $18\,{}^\circ\text{C}$ throughout the heating season. The table gives sample data on the mean outdoor temperatures of the months of the heating season, and a month has $720\,\text{h}$. Assume that the heating energy equals the energy conducted through the structures.

   Heating is provided either (A) by electric radiators, whose efficiency is practically $100\,\%$ and the price of electricity is $0.15\,\tfrac{\text{€}}{\text{kWh}}$, or (B) by a wood-burning fireplace, whose efficiency is $70\,\%$, the heating value of firewood is $15\,\tfrac{\text{MJ}}{\text{kg}}$ and the price is $0.18\,\tfrac{\text{€}}{\text{kg}}$.

   a) Calculate the heating energy needed over the whole heating season, in kilowatt-hours.

   b) Calculate the cost of each option over the heating season.

   c) Which option is cheaper, and how many euros does it save? Justify in writing which assumptions and prices affect the conclusion.

   Give the amount in euros from part c) in the answer box.

   Mean outdoor temperature of the month (sample data)

   Month	$T_\text{out}\,({}^\circ\text{C})$
   October	5
   November	0
   December	−4
   January	−9
   February	−7
   March	−3

   €Check
   Answer

   $480\,\text{€}$

   Show the solution

   Given

   average $U = 0.50\,\tfrac{\text{W}}{\text{m}^2\text{K}}$, $A = 120\,\text{m}^2$, $T_\text{in} = 18\,{}^\circ\text{C}$, $t = 720\,\text{h}$ per month; outdoor temperatures from the table: $5\,{}^\circ\text{C}$, $0\,{}^\circ\text{C}$, $-4\,{}^\circ\text{C}$, $-9\,{}^\circ\text{C}$, $-7\,{}^\circ\text{C}$, $-3\,{}^\circ\text{C}$; A: $\eta_\text{A} = 1.00$, $0.15\,\tfrac{\text{€}}{\text{kWh}}$; B: $\eta_\text{B} = 0.70$, heating value $H = 15\,\tfrac{\text{MJ}}{\text{kg}}$, $0.18\,\tfrac{\text{€}}{\text{kg}}$

   Find

   the saving in cost over the heating season from the cheaper option, in euros

   Principle

   The heating energy equals the energy conducted through the structures. The power of each month is $P = UA\Delta T$ and the energy is $P\,t$. Since $UA$, $t$ and the indoor temperature are the same, the energy of the season is $UA\,t\,\Sigma\Delta T$. The fuel has to supply this energy divided by the efficiency, and the mass of firewood is obtained from the heating value.

   Formula

   $E = UA\,t\,\Sigma\Delta T$, the cost of electricity

   $C_\text{A}$$= \dfrac{E}{\eta_\text{A}}\,h_\text{A}$

   and of wood

   $C_\text{B}$$= \dfrac{E}{\eta_\text{B}\,H}\,h_\text{B}$

   so the saving is $C_\text{A} - C_\text{B}$

   Conversions

   The temperature differences $\Delta T = 18\,{}^\circ\text{C} - T_\text{out}$ are the same in kelvins: $13$, $18$, $22$, $27$, $25$ and $21\,\text{K}$. The heat loss coefficient is

   $UA$$= 0.50 \cdot 120\,\tfrac{\text{W}}{\text{K}}$

   $= 60\,\tfrac{\text{W}}{\text{K}}$

   $= 0.060\,\tfrac{\text{kW}}{\text{K}}$

   The heating value $15\,\tfrac{\text{MJ}}{\text{kg}}$ is converted: $1\,\text{kWh} = 3.6\,\text{MJ}$, so

   $H$$= \dfrac{15}{3.6}\,\tfrac{\text{kWh}}{\text{kg}}$

   $= 4.167\,\tfrac{\text{kWh}}{\text{kg}}$

   Substitution

   The sum of the temperature differences is $13 + 18 + 22 + 27 + 25 + 21 = 126\,\text{K}$, so

   $E$$= 0.060\,\tfrac{\text{kW}}{\text{K}} \cdot 720\,\text{h} \cdot 126\,\text{K}$

   $= 5443\,\text{kWh}$

   Electricity:

   $C_\text{A}$$= \dfrac{5443\,\text{kWh}}{1.00} \cdot 0.15\,\tfrac{\text{€}}{\text{kWh}}$

   $= 816.5\,\text{€}$

   Wood: mass $= \dfrac{5443\,\text{kWh}}{0.70 \cdot 4.167\,\tfrac{\text{kWh}}{\text{kg}}} = 1866\,\text{kg}$, so

   $C_\text{B}$$= 1866\,\text{kg} \cdot 0.18\,\tfrac{\text{€}}{\text{kg}}$

   $= 335.9\,\text{€}$

   Saving:

   $C_\text{A} - C_\text{B}$$= 816.5\,\text{€} - 335.9\,\text{€}$

   $= 480.6\,\text{€}$

   Answer

   about $480\,\text{€}$ ($480.6\,\text{€}$): by this model, heating with wood is about $480\,\text{€}$ cheaper; two significant figures, because $U = 0.50$ and $A = 120\,\text{m}^2$ are known to two figures.

   Reasonableness

   The energy of the season, $5443\,\text{kWh} \approx 19.6\,\text{GJ}$, and the energy content of the wood,

   $1866\,\text{kg} \cdot 15\,\tfrac{\text{MJ}}{\text{kg}} \cdot 0.70$

   $\approx 19.6\,\text{GJ}$

   agree. The price of one useful kilowatt-hour is $0.15\,\tfrac{\text{€}}{\text{kWh}}$ for electricity and

   $\dfrac{0.18}{0.70 \cdot 4.167}\,\tfrac{\text{€}}{\text{kWh}}$

   $\approx 0.062\,\tfrac{\text{€}}{\text{kWh}}$

   for wood, so electricity is about $2.4$ times as expensive, and the cost ratio $816.5 : 335.9 \approx 2.4$ is the same.

   Model answer for part c)

   heating with wood is cheaper. The conclusion depends on the prices and efficiencies: if the price of wood were above about $0.44\,\tfrac{\text{€}}{\text{kg}}$, electricity would be cheaper. The model assumes a constant indoor temperature, the monthly mean temperatures, and that only $UA$ describes the losses; ventilation, the effort of using a fireplace and the storage of wood are not in the calculation. The prices and temperatures of the sample data are made up.

5. Exercise 16-20

   The ice sheet of an ice rink is $60\,\text{m}$ long and $30\,\text{m}$ wide. The rink is frozen by spraying a layer of water $0.030\,\text{m}$ thick, at a temperature of $12\,{}^\circ\text{C}$, onto a concrete slab. The layer is frozen completely, and the temperature of the finished ice is $-5.0\,{}^\circ\text{C}$. The density of water is $1000\,\tfrac{\text{kg}}{\text{m}^3}$, the specific heat capacity of water is $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and that of ice is $2100\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$. The refrigeration machine removes the energy at a constant power over $10$ hours. Assume that all of the power is used to cool the water, freeze it and cool the ice.

   a) Calculate the mass of water to be frozen.

   b) Calculate the energies of the three stages separately: cooling of the water to $0\,{}^\circ\text{C}$, freezing, and cooling of the ice to $-5.0\,{}^\circ\text{C}$.

   c) Calculate the average power the refrigeration machine needs, in kilowatts.

   d) Which stage requires the most energy, and why is the real cooling power needed greater than the calculated one?

   Give the result of part c) in the answer box.

   kWCheck
   Answer

   $590\,\text{kW}$

   Show the solution

   Given

   length $60\,\text{m}$, width $30\,\text{m}$, thickness of the water layer before freezing $0.030\,\text{m}$, initial temperature of the water $12\,{}^\circ\text{C}$, final temperature of the ice $-5.0\,{}^\circ\text{C}$, $\rho = 1000\,\tfrac{\text{kg}}{\text{m}^3}$, $c_\text{water} = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, $c_\text{ice} = 2100\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, time $t = 10\,\text{h}$; from the table $s = 334\,\tfrac{\text{kJ}}{\text{kg}}$

   Find

   average power $P$ of the refrigeration machine in kilowatts

   Principle

   The machine removes energy from the water in three stages, and all the power is used for them: $P\,t = Q_1 + Q_2 + Q_3$. The water first cools to $0\,{}^\circ\text{C}$ ($Q_1 = c_\text{water}\,m\,\Delta T_1$), then freezes at constant temperature ($Q_2 = s\,m$: when freezing, the same energy is released as is needed for melting), and finally the ice cools to $-5.0\,{}^\circ\text{C}$ ($Q_3 = c_\text{ice}\,m\,\Delta T_3$). The temperature does not change during freezing, so $c\,\Delta T$ is not used for it.

   Formula

   $m = \rho\,V$, $P\,t = c_\text{water}\,m\,\Delta T_1 + s\,m + c_\text{ice}\,m\,\Delta T_3$, so

   $P$$= \dfrac{m\,(c_\text{water}\,\Delta T_1 + s + c_\text{ice}\,\Delta T_3)}{t}$

   Conversions

   The changes in temperature are the same in kelvins: $\Delta T_1 = 12\,\text{K}$ and $\Delta T_3 = 5.0\,\text{K}$.

   $s$$= 334\,\tfrac{\text{kJ}}{\text{kg}}$

   $= 334\,000\,\tfrac{\text{J}}{\text{kg}}$

   $t = 10\,\text{h} = 36\,000\,\text{s}$.

   Substitution

   The volume is

   $V$$= 60\,\text{m} \cdot 30\,\text{m} \cdot 0.030\,\text{m}$

   $= 54\,\text{m}^3$

   and the mass is

   $m$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 54\,\text{m}^3$

   $= 54\,000\,\text{kg}$

   Cooling of the water:

   $Q_1$$= 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 54\,000\,\text{kg} \cdot 12\,\text{K}$

   $= 2.715 \cdot 10^{9}\,\text{J}$

   Freezing:

   $Q_2$$= 334\,000\,\tfrac{\text{J}}{\text{kg}} \cdot 54\,000\,\text{kg}$

   $= 18.04 \cdot 10^{9}\,\text{J}$

   Cooling of the ice:

   $Q_3$$= 2100\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 54\,000\,\text{kg} \cdot 5.0\,\text{K}$

   $= 0.567 \cdot 10^{9}\,\text{J}$

   In total $Q = 21.32 \cdot 10^{9}\,\text{J}$, and the power is

   $P$$= \dfrac{21.32 \cdot 10^{9}\,\text{J}}{36\,000\,\text{s}}$

   $= 592\,200\,\text{W}$

   $= 592\,\text{kW}$

   Answer

   $590\,\text{kW}$

   Two significant figures, because the thickness of the water layer $0.030\,\text{m}$ and the temperature of the ice $-5.0\,{}^\circ\text{C}$ have two figures.

   Reasonableness

   Estimate: freezing $54\,000\,\text{kg}$ needs about $0.33\,\text{MJ/kg} \cdot 54\,000\,\text{kg} \approx 18\,\text{GJ}$, and altogether about $21\,\text{GJ}$, which is $21.3\,\text{GJ} : 3.6\,\text{GJ/MWh} \approx 5.9\,\text{MWh}$, or about $5900\,\text{kWh}$; over ten hours the power is about $590\,\text{kW}$.

   Model answer for part d)

   freezing requires the most, about $85\,\%$ of the total energy, because the heat of freezing $s = 334\,\text{kJ/kg}$ corresponds roughly to a change in temperature of $80\,\text{K}$ in water ($334 : 4.19 \approx 80$), whereas the cooling of the water and the ice is only $12\,\text{K}$ and $5.0\,\text{K}$. The real cooling power needed is greater than the calculated one, because additional heat comes from the air of the hall, the lighting, the spectators, the concrete slab and the ground; the model therefore gives a lower limit. The electrical power of the machine also depends on its coefficient of performance.