Internal energy, heat and conservation of energy

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

1. Exercise 6-1Core

   Bread dough is kneaded by hand. While kneading, the hands do $450\,\text{J}$ of work on the dough, and a quantity of heat of $120\,\text{J}$ is transferred from the warm hands to the dough. The dough is the system.

   By how much does the internal energy of the dough change?

   JCheck
   Answer

   $570\,\text{J}$

   Show the solution

   Given

   the system is the dough; work done by the hands $W = +450\,\text{J}$ (the surroundings do work on the dough), heat transferred from the hands $Q = +120\,\text{J}$ (heat is transferred into the dough)

   Find

   the change in internal energy of the dough $\Delta U$

   Principle

   Energy is conserved: the internal energy of the system changes by exactly the amount of energy transferred into it as heat and work (the first law of thermodynamics). The law applies because the dough is a clearly bounded system and all energy crosses its boundary only as work or heat. Both transfers bring energy into the system, so both are positive.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: both $Q$ and $W$ are given in joules, and the result is wanted in joules.

   Substitution

   $\Delta U$$= 120\,\text{J} + 450\,\text{J}$

   $= 570\,\text{J}$

   Answer

   $570\,\text{J}$

   In whole joules, because both transfers are given in whole joules.

   Reasonableness

   Both transfers bring energy into the dough, so the internal energy must increase, and the result $570\,\text{J}$ is larger than either single transfer:

   $570\,\text{J}$$> 450\,\text{J}$

   $> 120\,\text{J}$

   Kneaded dough warms up slightly, which fits experience.

2. Exercise 6-2Core

   A cup of hot coffee cools on a table at a cottage. A quantity of heat of $8.4\,\text{kJ}$ is transferred from the coffee to the surroundings. No work is done on the coffee, and the coffee does no work.

   By how much does the internal energy of the coffee change? Give the answer with its sign in kilojoules.

   kJCheck
   Answer

   $-8.4\,\text{kJ}$

   Show the solution

   Given

   the system is the coffee; $Q = -8.4\,\text{kJ}$ (heat is transferred out of the coffee), $W = 0$

   Find

   the change in internal energy of the coffee $\Delta U$ in kilojoules

   Principle

   Energy is conserved, so the change in internal energy is the sum of the heat and work transferred into the system. Heat leaves the coffee, so $Q$ is negative. No work is done in either direction, so $W = 0$ and only heat changes the internal energy.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: $Q$ is given in kilojoules and the answer is wanted in kilojoules, so $W = 0\,\text{kJ}$ is enough.

   Substitution

   $\Delta U$$= -8.4\,\text{kJ} + 0\,\text{kJ}$

   $= -8.4\,\text{kJ}$

   Answer

   $-8.4\,\text{kJ}$

   Two significant figures, like the quantity of heat $8.4\,\text{kJ}$.

   Reasonableness

   The sign is negative, because the coffee gives off energy and cools down. Energy is not lost: the same $8.4\,\text{kJ}$ is transferred to the table and the room air, so the internal energy of the surroundings increases by $+8.4\,\text{kJ}$ and the sum for the whole is zero.

3. Exercise 6-3

   A party balloon is taken into a warm sauna. A quantity of heat of $350\,\text{J}$ is transferred to the air in the balloon, and at the same time the balloon expands, so the air does $100\,\text{J}$ of work on the surroundings.

   By how much does the internal energy of the air in the balloon change?

   JCheck
   Answer

   $250\,\text{J}$

   Show the solution

   Given

   the system is the air in the balloon; $Q = +350\,\text{J}$ (heat is transferred into the air), $W = -100\,\text{J}$ (the air does work on the surroundings as it expands)

   Find

   the change in internal energy of the air $\Delta U$

   Principle

   Energy is conserved: the change in internal energy is the sum of the heat and work transferred into the system. The air gains energy as heat but gives up part of it as expansion work, so the work is negative and reduces the increase in internal energy.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: both transfers are given in joules.

   Substitution

   $\Delta U$$= 350\,\text{J} + (-100\,\text{J})$

   $= 250\,\text{J}$

   Answer

   $250\,\text{J}$

   In whole joules, because the transfers are given in whole joules.

   Reasonableness

   Part of the incoming energy goes into the expansion work, so the internal energy increases by less than $350\,\text{J}$: $250\,\text{J} < 350\,\text{J}$. The internal energy still increases, so the temperature of the air rises. Energy balance: $350\,\text{J}$ came in, $100\,\text{J}$ went out, $250\,\text{J}$ remained.

4. Exercise 6-4

   In the cylinder of a car engine the fuel has burnt, and the hot combustion gas expands, pushing the piston. The gas does $320\,\text{J}$ of work on the piston, and at the same time a quantity of heat of $180\,\text{J}$ is transferred from the gas to the cylinder walls.

   By how much does the internal energy of the gas change? Give the answer with its sign.

   JCheck
   Answer

   $-500\,\text{J}$

   Show the solution

   Given

   the system is the combustion gas; $W = -320\,\text{J}$ (the gas does work on the piston), $Q = -180\,\text{J}$ (heat is transferred from the gas to the walls)

   Find

   the change in internal energy of the gas $\Delta U$ with its sign

   Principle

   Energy is conserved, so the change in internal energy is the sum of the heat and work. Here the gas gives up energy by two routes: as work when it expands and as heat to the walls. Both are out of the system, so both are negative.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: both transfers are given in joules.

   Substitution

   $\Delta U$$= -180\,\text{J} + (-320\,\text{J})$

   $= -500\,\text{J}$

   Answer

   $-500\,\text{J}$

   In whole joules, because the transfers are given in whole joules.

   Reasonableness

   Energy leaves the gas by two routes and none comes in, so the internal energy must decrease and the gas must cool. The decrease equals the sum of the transfers out: $180\,\text{J} + 320\,\text{J} = 500\,\text{J}$.

5. Exercise 6-5

   Air is let out of a bicycle tyre through the valve. The escaping air expands and does $85\,\text{J}$ of work on the surroundings. At the same time a quantity of heat of $15\,\text{J}$ has time to be transferred to it from the surroundings.

   By how much does the internal energy of the escaping air change? Give the answer with its sign.

   JCheck
   Answer

   $-70\,\text{J}$

   Show the solution

   Given

   the system is the escaping air; $W = -85\,\text{J}$ (the air does work on the surroundings), $Q = +15\,\text{J}$ (heat is transferred into the air)

   Find

   the change in internal energy of the air $\Delta U$ with its sign

   Principle

   Energy is conserved, so the change in internal energy is the sum of the heat and work. The air expands and does work, so $W$ is negative. Only a little heat has time to be transferred from the surroundings, because the release is fast.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: both transfers are given in joules.

   Substitution

   $\Delta U$$= 15\,\text{J} + (-85\,\text{J})$

   $= -70\,\text{J}$

   Answer

   $-70\,\text{J}$

   In whole joules, because both transfers are given in whole joules.

   Reasonableness

   The expansion work takes away more energy than the heat brings in ($85\,\text{J} > 15\,\text{J}$), so the internal energy must decrease and the air must cool. This is why the valve feels cold when air is let out.

6. Exercise 6-6

   A coffee cup and a big pot are both full of water, and the temperature of the water in both is $60\,{}^\circ\text{C}$. The pot holds ten times as much water as the cup.

   Which statement about the internal energy of the waters is correct?

   Choose an option.

   • The pot has the greater internal energy, because at the same temperature there are more particles.
   • Both have the same internal energy, because the temperature of the water is the same in both.
   • The pot has more heat, because heat is stored in water in proportion to the amount of water.
   • The cup has the greater internal energy, because a small amount of water heats up faster than a large one.

   Answer
   I have done my solution.
   Answer

   The pot has the greater internal energy, because at the same temperature there are more particles.

   Show the solution

   Find

   which amount of water has the greater internal energy when the temperature is the same

   Principle

   Internal energy is the sum of the kinetic energy of the particles of the system and the energy of the interactions between the particles. Temperature tells how vigorously the particles move, but internal energy also depends on the number of particles. Heat is not an amount stored in a body, but the transfer of energy because of a temperature difference.

   Model answer

   The pot has the greater internal energy, because at the same temperature there are more particles. The claim that the internal energy is the same mixes up temperature and internal energy. The cup cannot have the greater internal energy, because it has fewer particles. The claim that the pot has more heat is wrong, because what is stored in a body is internal energy, not heat.

7. Exercise 6-7

   Hot and cold water are poured into an insulated container, and the water is stirred. No energy is transferred out of the container, the work done by the stirring is negligible, and the warming of the container itself can be ignored.

   Which statement about energy is correct?

   Choose an option.

   • The internal energy of the hot water decreases by as much as that of the cold water increases, so energy is conserved.
   • The cold water gains more energy than the hot water gives off, because there is room in cold water.
   • The internal energy of each water stays the same, and only the temperatures even out with each other.
   • The hot water gives off heat, but part of it is lost in the stirring, and the total energy decreases.

   Answer
   I have done my solution.
   Answer

   The internal energy of the hot water decreases by as much as that of the cold water increases, so energy is conserved.

   Show the solution

   Find

   what happens to energy when hot and cold water are mixed in an insulated container (the warming of the container itself is ignored)

   Principle

   The total energy of an isolated system stays constant (conservation of energy). Heat is transferred from the hotter to the colder until the temperatures are the same. So the energy given off by the hot water is exactly the energy gained by the cold water.

   Model answer

   The internal energy of the hot water decreases by as much as the internal energy of the cold water increases, so the total energy is conserved. The claim that energy is lost in the stirring breaks conservation of energy, because no energy is transferred out of the container. The internal energy cannot stay the same in both, because the temperatures change. Nor can the cold water gain more energy than the hot water gives off.

8. Exercise 6-8

   An immersion heater transfers heat to the water in a bucket at a steady power of $0.80\,\text{kW}$ for $45\,\text{s}$. The bucket is well insulated, so no heat escapes to the surroundings, and no work is done on the water.

   By how much does the internal energy of the water increase? Give the answer in kilojoules.

   kJCheck
   Answer

   $36\,\text{kJ}$

   Show the solution

   Given

   the system is the water; power $P = 0.80\,\text{kW}$, time $t = 45\,\text{s}$, no heat escapes to the surroundings, $W = 0$

   Find

   the increase in internal energy of the water $\Delta U$ in kilojoules

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The heater transfers energy to the water as heat at a steady power, so the quantity of heat transferred to the water is the product of power and time. No work is done, so $W = 0$.

   Formula

   $Q = Pt$ and $\Delta U = Q + W$, so $\Delta U = Pt$

   Conversions

   $P = 0.80\,\text{kW} = 800\,\text{W}$, so that $Pt$ comes out in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $\Delta U$$= 800\,\text{W} \cdot 45\,\text{s}$

   $= 36\,000\,\text{J}$

   $= 36\,\text{kJ}$

   Answer

   $36\,\text{kJ}$

   Two significant figures, like the data $0.80\,\text{kW}$ and $45\,\text{s}$.

   Reasonableness

   The heater transfers $800\,\text{J}$ every second, and in a whole minute it would transfer $0.80\,\text{kW} \cdot 60\,\text{s} = 48\,\text{kJ}$. The time $45\,\text{s}$ is shorter than a minute, so the result $36\,\text{kJ}$ is smaller: $36\,\text{kJ} < 48\,\text{kJ}$. The result is positive, because energy comes into the water and none leaves it.

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

1. Exercise 6-9Core

   The piston of a compressor compresses the air in a cylinder. The graph shows the force $F$ of the piston on the air as a function of the distance $s$ the piston moves during the compression. The internal energy of the air increases by $45\,\text{J}$ in the compression.

   How large a quantity of heat $Q$ is transferred into the air? Read the values you need from the graph and give the answer with its sign: a negative $Q$ means that heat is transferred out of the air.

   JCheck
   Answer

   $-15\,\text{J}$

   Show the solution

   Given

   the system is the air; the force of the piston on the air is given as a function of distance in the graph, $\Delta U = +45\,\text{J}$

   Find

   the quantity of heat $Q$ transferred into the air, with its sign

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The force changes along the distance, so the work is not calculated as the product $Fs$ but as the area between the $(s,F)$ graph and the horizontal axis. The piston is part of the surroundings, which do work on the air, so $W$ is positive. The heat is found as the difference between the change in internal energy and the work.

   Reading the graph

   The line starts at the grid intersection $(0\,\text{m}$; $200\,\text{N})$ and ends at the intersection $(0.20\,\text{m}$; $400\,\text{N})$. The area is a trapezoid whose parallel sides are $F_1 = 200\,\text{N}$ and $F_2 = 400\,\text{N}$ and whose height is $s = 0.20\,\text{m}$.

   Formula

   $W = \dfrac{F_1 + F_2}{2}\, s$ and $\Delta U = Q + W$, so $Q = \Delta U - W$

   Conversions

   None needed: the force is in newtons and the distance in metres, so $1\,\text{N}\cdot\text{m} = 1\,\text{J}$.

   Substitution

   $W$$= \dfrac{200\,\text{N} + 400\,\text{N}}{2} \cdot 0.20\,\text{m}$

   $= 300\,\text{N} \cdot 0.20\,\text{m}$

   $= 60\,\text{J}$

   so

   $Q$$= 45\,\text{J} - 60\,\text{J}$

   $= -15\,\text{J}$

   Answer

   $-15\,\text{J}$

   Two significant figures, like the data $0.20\,\text{m}$ and $45\,\text{J}$. The values of the graph are read accurately from the grid, so $-14.5\ldots-15.5\,\text{J}$ is accepted.

   Reasonableness

   The work brought $60\,\text{J}$ into the air, but the internal energy increased by only $45\,\text{J}$, so $15\,\text{J}$ must have left as heat and $Q$ is negative. This fits the fact that compressed air is warm and gives off heat to the cylinder walls. Balance: $-15\,\text{J} + 60\,\text{J} = 45\,\text{J}$.

2. Exercise 6-10Core

   The heating element of a kettle transfers energy to the water as heat at a steady power of $2.0\,\text{kW}$ for 90 seconds. At the same time, a quantity of heat of $20\,\text{kJ}$ is transferred from the water through the casing of the kettle to the surroundings.

   By how much does the internal energy of the water increase? Give the answer in kilojoules.

   kJCheck
   Answer

   $160\,\text{kJ}$

   Show the solution

   Given

   the system is the water; power of the heating element $P = 2.0\,\text{kW}$, time $t = 90\,\text{s}$, heat transferred through the casing to the surroundings $20\,\text{kJ}$; no work is done, so $W = 0$

   Find

   the increase in internal energy of the water $\Delta U$ in kilojoules

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The heating element transfers energy to the water as heat, and part of it escapes. The net heat transferred to the water is the energy coming in minus the energy going out, and since the water does not expand and no work is done on it, $W = 0$.

   Formula

   energy transferred by the heating element $Q_\text{in} = Pt$, net heat $Q = Pt - Q_\text{out}$ and $\Delta U = Q + W$, so $\Delta U = Pt - Q_\text{out}$

   Conversions

   $P = 2.0\,\text{kW} = 2000\,\text{W}$, so that $Pt$ comes out in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $Pt$$= 2000\,\text{W} \cdot 90\,\text{s}$

   $= 180\,000\,\text{J}$

   $= 180\,\text{kJ}$

   So

   $\Delta U$$= 180\,\text{kJ} - 20\,\text{kJ} + 0$

   $= 160\,\text{kJ}$

   Answer

   $160\,\text{kJ}$

   Two significant figures: the difference $180\,\text{kJ} - 20\,\text{kJ}$ is known to within ten kilojoules, because the heat $20\,\text{kJ}$ is given only that precisely.

   Reasonableness

   About $\tfrac{160}{180} \approx 89\,\%$ of the energy from the heating element stays in the water and only a small part escapes through the casing, which fits a well-insulated kettle. Roughly: $2\,\text{kW}$ for two minutes would be $240\,\text{kJ}$, so $90\,\text{s}$ gives about three quarters of that, or $180\,\text{kJ}$.

3. Exercise 6-11

   In the cylinder of the engine of a steamship sailing on Lake Saimaa, steam expands and does $2.0\,\text{kJ}$ of work on the piston. At the same time, the internal energy of the steam decreases by $1.2\,\text{kJ}$.

   How large a quantity of heat is transferred into the steam? Give the answer in joules.

   JCheck
   Answer

   $800\,\text{J} = 0.8\,\text{kJ}$

   Show the solution

   Given

   the system is the steam; $W = -2.0\,\text{kJ}$ (the steam does work), $\Delta U = -1.2\,\text{kJ}$

   Find

   the quantity of heat $Q$ transferred into the steam, in joules

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The change in internal energy and the work are known, so the heat is found as their difference. The steam expands and does work, so $W$ is negative.

   Formula

   $\Delta U = Q + W$, so $Q = \Delta U - W$

   Conversions

   The answer is wanted in joules:

   $\Delta U$$= -1.2\,\text{kJ}$

   $= -1200\,\text{J}$

   $W = -2.0\,\text{kJ} = -2000\,\text{J}$.

   Substitution

   $Q$$= -1200\,\text{J} - (-2000\,\text{J})$

   $= 800\,\text{J}$

   Answer

   $800\,\text{J} = 0.8\,\text{kJ}$

   ; the difference is known to within a hundred joules, like the data, so one significant figure.

   Reasonableness

   $Q$ is positive, so heat is transferred into the steam. The steam gave up $2000\,\text{J}$ as work, but its internal energy decreased by only $1200\,\text{J}$, so the missing $800\,\text{J}$ must have come as heat from the hot cylinder walls. Balance: $800\,\text{J} - 2000\,\text{J} = -1200\,\text{J}$.

4. Exercise 6-12

   A smoothie made of frozen blueberries is blended in a blender. The blade does work on the smoothie at a power of $250\,\text{W}$. The smoothie is colder than the air in the kitchen, and during blending a quantity of heat of $2.0\,\text{kJ}$ is transferred to it through the walls of the jug. After blending, the internal energy of the smoothie has increased by $12\,\text{kJ}$.

   How long was the blender on?

   sCheck
   Answer

   $40\,\text{s}$

   Show the solution

   Given

   the system is the smoothie; power of the blade $P = 250\,\text{W}$, $Q = +2.0\,\text{kJ}$ (heat is transferred into the smoothie), $\Delta U = +12\,\text{kJ}$

   Find

   the blending time $t$

   Principle

   Energy is conserved, so $\Delta U = Q + W$. Part of the increase in internal energy comes as heat and the rest as work done by the blade. The work of the blade is the product of power and time, because the power is constant: $W = Pt$.

   Formula

   $\Delta U = Q + W$, so $W = \Delta U - Q$. Since $P = \dfrac{W}{t}$, we get

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

   $= \dfrac{\Delta U - Q}{P}$

   Conversions

   Energies from kilojoules to joules, so that the result comes out in seconds: $1\,\text{kJ} = 1000\,\text{J}$, so $\Delta U = 12\,000\,\text{J}$ and $Q = 2000\,\text{J}$. The unit of power is $1\,\text{W} = 1\,\tfrac{\text{J}}{\text{s}}$.

   Substitution

   $W$$= 12\,\text{kJ} - 2.0\,\text{kJ}$

   $= 10\,\text{kJ}$

   $= 10\,000\,\text{J}$

   So

   $t$$= \dfrac{10\,000\,\text{J}}{250\,\text{W}}$

   $= 40\,\text{s}$

   Answer

   $40\,\text{s}$

   Two significant figures, because the work $W = 12\,\text{kJ} - 2.0\,\text{kJ} = 10\,\text{kJ}$ is known to a precision of $1\,\text{kJ}$ (the internal energy $12\,\text{kJ}$ is given in whole kilojoules).

   Reasonableness

   Blending for under a minute is a normal time for a blender. Check with the power: $250\,\text{W} \cdot 40\,\text{s} = 10\,000\,\text{J}$, and $10\,\text{kJ} + 2.0\,\text{kJ} = 12\,\text{kJ}$. If no heat were transferred, it would take $\tfrac{12\,000}{250} = 48\,\text{s}$, so the result is shorter than that, as it should be.

5. Exercise 6-13

   A gas in a closed cylinder goes through a cycle of four steps and returns after the fourth step exactly to its initial state: the amount of gas, the temperature and the other state variables are the same as at the start.

   The table gives, as sample data, the quantity of heat $Q$ transferred to the gas and the work $W$ done on the gas in each step; the work in step 4 is not known.

   How large is the work $W$ in step 4? Give the answer with its sign: a negative $W$ means that the gas does work on the surroundings.

   Heat and work of the gas in the steps of the cycle (sample data)

   Step	$Q\,(\text{J})$	$W\,(\text{J})$
   1	600	−250
   2	−150	0
   3	−320	210
   4	0	?

   JCheck
   Answer

   $-90\,\text{J}$

   Show the solution

   Given

   the system is the gas; steps 1–3: $Q_1 = +600\,\text{J}$, $W_1 = -250\,\text{J}$; $Q_2 = -150\,\text{J}$, $W_2 = 0$; $Q_3 = -320\,\text{J}$, $W_3 = +210\,\text{J}$; step 4: $Q_4 = 0$, $W_4$ unknown; the gas returns to its initial state at the end of the cycle

   Find

   the work $W_4$ in step 4 with its sign

   Principle

   Energy is conserved, so $\Delta U = Q + W$ for the whole cycle. The gas returns exactly to its initial state, so its internal energy is the same as at the start. Internal energy is a function of state, so in a cycle its change is zero: $\Delta U = 0$. Then the sum of the heats and works of the cycle is zero.

   Formula

   $\Delta U$$= (Q_1 + Q_2 + Q_3 + Q_4) + (W_1 + W_2 + W_3 + W_4)$

   $= 0$

   so $W_4 = -(Q_1 + Q_2 + Q_3 + Q_4) - (W_1 + W_2 + W_3)$

   Conversions

   None needed: all energies are given in joules.

   Substitution

   The sum of the heats is $600\,\text{J} - 150\,\text{J} - 320\,\text{J} + 0 = 130\,\text{J}$ and the sum of the known works is $-250\,\text{J} + 0 + 210\,\text{J} = -40\,\text{J}$. So

   $W_4$$= -130\,\text{J} - (-40\,\text{J})$

   $= -90\,\text{J}$

   Answer

   $-90\,\text{J}$

   In whole joules, because the data are given in whole joules. A negative work means that the gas does $90\,\text{J}$ of work on the surroundings.

   Reasonableness

   In the cycle heat of $130\,\text{J}$ in total comes into the gas, so it must do $130\,\text{J}$ of work on the surroundings in total: $-250\,\text{J} + 0 + 210\,\text{J} - 90\,\text{J} = -130\,\text{J}$, and $130\,\text{J} - 130\,\text{J} = 0$. In step 4 no heat is transferred, so doing work decreases the internal energy of the gas:

   $\Delta U_4$$= 0 + (-90\,\text{J})$

   $= -90\,\text{J}$

   and the gas cools.

6. Exercise 6-14

   A student solved a problem and got the result $\Delta U = 40\,\text{J}$. The problem: the air in the cylinder of a bicycle pump is compressed, so that the piston does $30\,\text{J}$ of work on the air, and at the same time a quantity of heat of $10\,\text{J}$ is transferred from the air through the walls of the cylinder to the surroundings. The student's solution: “Work is done on the air, so $W = +30\,\text{J}$. The quantity of heat is $10\,\text{J}$, so $Q = 10\,\text{J}$. So $\Delta U = Q + W = 10\,\text{J} + 30\,\text{J} = 40\,\text{J}$.”

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

   b) Calculate the change in the internal energy of the air correctly.

   The answer box takes the result of part b) with its sign.

   JCheck
   Answer

   $20\,\text{J}$

   Show the solution

   Given

   the system is the air in the pump; $W = +30\,\text{J}$ (the piston does work on the air), a quantity of heat of $10\,\text{J}$ is transferred out of the air; the student's result $\Delta U = 40\,\text{J}$

   Find

   a) the student's error, b) the correct $\Delta U$ with its sign

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The signs show the direction: energy into the system is positive and energy out of the system is negative. Heat leaving the air decreases the internal energy, so $Q$ is negative.

   Formula

   $\Delta U = Q + W$, and the quantity to find $\Delta U$ is already on the left-hand side of the equation

   Conversions

   None needed: both transfers are given in joules.

   Substitution

   a) The student gave the heat a plus sign, although heat is transferred out of the air and the correct value is $Q = -10\,\text{J}$. The sign of the work, $W = +30\,\text{J}$, is correct. b)

   $\Delta U$$= -10\,\text{J} + 30\,\text{J}$

   $= 20\,\text{J}$

   Answer

   $20\,\text{J}$

   In whole joules, because the transfers are given in whole joules.

   Reasonableness

   The work brought $30\,\text{J}$ into the air and $10\,\text{J}$ left as heat, so the internal energy must increase by less than the work: $20\,\text{J} < 30\,\text{J}$. The student's result was larger than the work, although energy left the air, which cannot be right.

7. Exercise 6-15

   Students pull a wooden block along a table at a steady speed with a force meter. The block and the table form the system.

   The table gives, as sample data, the pulling force $F$ in five pulls; in each pull the block moves a distance of $0.600\,\text{m}$. The work done by the pulling force is changed into internal energy by friction. Assume that no heat is transferred to the surroundings, and use the mean of the measurements as the pulling force.

   By how much does the internal energy of the system increase during 20 pulls? Justify briefly why the real increase is probably slightly smaller than the calculated one.

   Pulling force on the block in five pulls (sample data)

   Pull	$F\,(\text{N})$
   1	4.1
   2	4.3
   3	4.2
   4	4.0
   5	4.4

   JCheck
   Answer

   $50.4\,\text{J}$

   Show the solution

   Given

   the system is the block and the table; number of pulls $n = 20$, distance per pull $s = 0.600\,\text{m}$, forces $4.1\,\text{N}$, $4.3\,\text{N}$, $4.2\,\text{N}$, $4.0\,\text{N}$ and $4.4\,\text{N}$; $Q = 0$

   Find

   the increase in internal energy $\Delta U$ during 20 pulls, and why the real increase is smaller

   Principle

   Energy is conserved, so $\Delta U = Q + W$. By assumption $Q = 0$, so $\Delta U = W$. The pulling force does work $W = Fs$ on the system, and friction changes it into internal energy. The measurements vary slightly, so the mean is used as the pulling force.

   Formula

   $F_\text{mean}$$= \dfrac{\text{sum of the measurements}}{5}$

   the work of one pull $W_1 = F_\text{mean}\, s$ and for all the pulls $\Delta U = nW_1$

   Conversions

   None needed: the force is in newtons and the distance in metres, so the product is in joules.

   Substitution

   The sum of the measurements is

   $4.1\,\text{N} + 4.3\,\text{N} + 4.2\,\text{N} + 4.0\,\text{N} + 4.4\,\text{N}$

   $= 21.0\,\text{N}$

   so

   $F_\text{mean}$$= \dfrac{21.0\,\text{N}}{5}$

   $= 4.20\,\text{N}$

   The work of one pull is

   $W_1$$= 4.20\,\text{N} \cdot 0.600\,\text{m}$

   $= 2.52\,\text{J}$

   so

   $\Delta U$$= 20 \cdot 2.52\,\text{J}$

   $= 50.4\,\text{J}$

   Answer

   $50.4\,\text{J}$

   That is, about $50\,\text{J}$: the measurements are given to two significant figures and they vary by about $\pm 0.2\,\text{N}$ ($\approx 5\,\%$), so a result with two significant figures is honest. The real increase is probably slightly smaller, because the warmed block and table transfer some heat to the air, so $Q < 0$ and $\Delta U = Q + W < W$.

   Reasonableness

   The $2.52\,\text{J}$ of one pull is small, which is why the warming caused by friction is hard to notice in everyday life, but the $50.4\,\text{J}$ of $20$ pulls is already clearly measurable. Check: $20 \cdot 0.600\,\text{m} = 12.0\,\text{m}$ of pulling distance multiplied by the force $4.20\,\text{N}$ gives $50.4\,\text{J}$.

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

1. Exercise 6-16

   The nozzle of a bicycle pump is blocked with a finger, so no air can escape. The piston is pushed very slowly, and it pushes the air a distance of $0.15\,\text{m}$ with an average force of $120\,\text{N}$. The temperature of the air in the pump stays at room temperature throughout, and the air can be treated as an ideal gas.

   How large a quantity of heat $Q$ is transferred into the air? Give the answer with its sign and justify why the internal energy of the air does not change.

   JCheck
   Answer

   $-18\,\text{J}$

   Show the solution

   Given

   the system is the air in the pump (the amount does not change); $F = 120\,\text{N}$, $s = 0.15\,\text{m}$, the temperature stays constant

   Find

   the quantity of heat $Q$ transferred into the air, with its sign, and why $\Delta U$ does not change

   Principle

   Energy is conserved, so $\Delta U = Q + W$. According to the problem, the temperature stays constant and the amount of air does not change, so the kinetic energy of the particles and their number are unchanged, and because the air can be treated as an ideal gas, its internal energy does not change: $\Delta U = 0$. Then the energy brought in by the work must be transferred out as heat: $Q = -W$.

   Formula

   $W = Fs$ and $\Delta U = Q + W = 0$, so $Q = -W = -Fs$

   Conversions

   None needed: the force is in newtons and the distance in metres, so the product is in joules.

   Substitution

   $W$$= 120\,\text{N} \cdot 0.15\,\text{m}$

   $= 18\,\text{J}$

   (work is done on the air, so $W > 0$). So $Q = -18\,\text{J}$

   Answer

   $-18\,\text{J}$

   Two significant figures, like the data $0.15\,\text{m}$ and $120\,\text{N}$.

   Reasonableness

   Because the piston is pushed slowly, the energy brought in by the work has time to be transferred as heat through the pump walls to the room, and the temperature stays constant. Balance: $-18\,\text{J} + 18\,\text{J} = 0$. Limiting case: if the piston were pushed quickly, there would be no time for heat to be transferred, $Q \approx 0$, and the pump would get hot.

2. Exercise 6-17

   In a physics lesson at upper secondary school, a measurement like Joule's experiment is done.

   A $5.0\,\text{kg}$ weight hanging on a string descends slowly at a constant speed and turns a paddle wheel in water. The water container is very well insulated. The weight is always lifted back up and released, 12 times in total.

   From what height must the weight descend each time so that the internal energy of the water increases by $1.0\,\text{kJ}$ in total? Assume that the energy lost by the weight is transferred to the water as work done by the paddle wheel.

   mCheck
   Answer

   $1.7\,\text{m}$

   Show the solution

   Given

   the system is the water; mass of the weight $m = 5.0\,\text{kg}$, number of descents $n = 12$, $\Delta U = 1.0\,\text{kJ}$, $g = 9.81\,\text{m/s}^2$, insulated container

   Find

   the height $h$ of one descent

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The container is insulated, so no heat is transferred and $Q = 0$. The weight descends at a constant speed, so the rope pulls it with a force $F = G = mg$, and when it descends a height $h$ the rope and the paddle wheel do work $W_1 = Fh = mgh$ on the water. Then $\Delta U = W$.

   Formula

   work of one descent $W_1 = mgh$, so $\Delta U = nmgh$ and $h = \dfrac{\Delta U}{nmg}$

   Conversions

   $\Delta U$$= 1.0\,\text{kJ}$

   $= 1000\,\text{J}$

   Finally $1\,\text{J} = 1\,\text{N}\cdot\text{m}$ and $1\,\text{N} = 1\,\tfrac{\text{kg}\cdot\text{m}}{\text{s}^2}$, so the unit of the quotient is the metre.

   Substitution

   $h$$= \dfrac{1000\,\text{J}}{12 \cdot 5.0\,\text{kg} \cdot 9.81\,\text{m/s}^2}$

   $= \dfrac{1000\,\text{J}}{588.6\,\text{N}}$

   $= 1.6989\,\text{m}$

   Answer

   $1.7\,\text{m}$

   Two significant figures, like the mass $5.0\,\text{kg}$ and the energy $1.0\,\text{kJ}$.

   Reasonableness

   A drop of about 1.7 metres fits in a classroom. The work of one descent is

   $5.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.7\,\text{m}$

   $\approx 83\,\text{J}$

   and $12 \cdot 83\,\text{J} \approx 1000\,\text{J}$. Joule's experiment needs many repetitions, because one kilogram of water would warm up by only $\tfrac{1000\,\text{J}}{4190\,\text{J/K}} \approx 0.24\,\text{K}$.

3. Exercise 6-18

   A gas in a cylinder is taken in two different ways from the same initial temperature to the same final temperature; the amount of gas does not change. The gas can be treated as an ideal gas, in which the energy of the interactions between the particles is negligibly small.

   In way A a quantity of heat of $400\,\text{J}$ is transferred to the gas, and the gas does $150\,\text{J}$ of work on the surroundings. In way B the gas does $60\,\text{J}$ of work on the surroundings.

   How large a quantity of heat is transferred to the gas in way B? Justify why the change in the internal energy of the gas is the same in both ways.

   JCheck
   Answer

   $310\,\text{J}$

   Show the solution

   Given

   the system is the gas; way A: $Q_\text{A} = +400\,\text{J}$, $W_\text{A} = -150\,\text{J}$ (the gas does work); way B: $W_\text{B} = -60\,\text{J}$; same initial and final temperature, same amount of gas

   Find

   the quantity of heat $Q_\text{B}$, and why $\Delta U$ is the same in both ways

   Principle

   Energy is conserved, so $\Delta U = Q + W$ for both ways. According to the problem, the temperature and the amount of gas are the same in the initial and final states, so the kinetic energy of the particles is the same, and because the energy of the interactions is negligible in an ideal gas, the internal energy is the same too. So $\Delta U$ is the same in both ways, even though the heat and work differ: they depend on the path, while the internal energy depends only on the initial and final states.

   Formula

   $\Delta U_\text{A} = Q_\text{A} + W_\text{A}$ and $\Delta U_\text{B} = \Delta U_\text{A}$, so

   $Q_\text{B}$$= \Delta U_\text{A} - W_\text{B}$

   $= Q_\text{A} + W_\text{A} - W_\text{B}$

   Conversions

   None needed: all energies are given in joules.

   Substitution

   $\Delta U_\text{A}$$= 400\,\text{J} + (-150\,\text{J})$

   $= 250\,\text{J}$

   So

   $Q_\text{B}$$= 250\,\text{J} - (-60\,\text{J})$

   $= 310\,\text{J}$

   Answer

   $310\,\text{J}$

   ; in the addition the result is as precise as the data, which are given in whole joules.

   Reasonableness

   In way B the gas does less work, so the same increase in internal energy $250\,\text{J}$ needs less heat than in way A: $310\,\text{J} < 400\,\text{J}$. Balance:

   $310\,\text{J} + (-60\,\text{J})$

   $= 250\,\text{J}$

   $= \Delta U_\text{A}$
4. Exercise 6-19

   The piston in the cylinder of a bicycle pump is pushed quickly a distance of $0.20\,\text{m}$, and the air in the pump is compressed. The table gives, as sample data, the force $F$ of the piston on the air as a function of the distance $s$ moved by the piston; the force grows because the compressed air resists more and more. In a quick compression there is no time for heat to be transferred to the surroundings, so $Q = 0$.

   a) Calculate the work $W$ done by the piston on the air with the trapezoid rule: the intervals are of equal length, and the work is the area under the graph of force against distance.

   b) By how much does the internal energy of the air increase in the quick compression?

   c) The same piston is now compressed slowly to the same final position so that the temperature of the air stays at room temperature; the air can be treated as an ideal gas. In the slow compression the piston does $14.0\,\text{J}$ of work on the air. How large a quantity of heat $Q$ is transferred to the air, and why does the temperature of the air rise only in the quick compression?

   The answer box takes the result of part b).

   Force of the pump piston as a function of distance (sample data)

   $s\,(\text{m})$	$F\,(\text{N})$
   0.00	50
   0.04	62
   0.08	78
   0.12	98
   0.16	125
   0.20	160

   JCheck
   Answer

   $W \approx 18.7\,\text{J}$

   Show the solution

   Given

   the system is the air in the pump; $(s,F)$ table, $s = 0$ … $0.20\,\text{m}$ in steps of $0.04\,\text{m}$; in the quick compression $Q = 0$; in the slow compression the temperature is constant and $W_\text{slow} = 14.0\,\text{J}$

   Find

   a) the work $W$ with the trapezoid rule, b) the increase in internal energy $\Delta U$ in the quick compression, c) the quantity of heat $Q$ in the slow compression and an explanation

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The force changes along the distance, so the work is the area under the $(s,F)$ graph, which the trapezoid rule estimates. The piston does work on the air, so $W > 0$. In the quick compression there is no time for heat to be transferred, so $\Delta U = W$. In the slow compression the temperature and the amount of air stay the same, and because the air can be treated as an ideal gas, $\Delta U = 0$ and $Q = -W$.

   Reading the graph

   The table has five intervals, each of width $h = 0.04\,\text{m}$. The forces at the ends of the intervals are $50$, $62$, $78$, $98$, $125$ and $160\,\text{N}$. The area of each interval is a trapezoid whose parallel sides are the forces at the ends of the interval.

   Formula

   trapezoid rule: the work is the sum of the areas $h\,\dfrac{F_i + F_{i+1}}{2}$ of all the intervals, where $F_i$ and $F_{i+1}$ are the forces at the ends of the interval. Also $\Delta U_\text{quick} = Q + W = W$ and $Q_\text{slow} = -W_\text{slow}$

   Conversions

   None needed: the force is in newtons and the distance in metres, so $1\,\text{N}\cdot\text{m} = 1\,\text{J}$.

   Substitution

   Each force has two adjacent intervals around it, except at the ends; that is why half of each end force is taken. The sum is

   $25\,\text{N} + 62\,\text{N} + 78\,\text{N} + 98\,\text{N} + 125\,\text{N} + 80\,\text{N}$

   $= 468\,\text{N}$

   So

   $W$$= 0.04\,\text{m} \cdot 468\,\text{N}$

   $= 18.72\,\text{J}$

   and

   $\Delta U$$= 0 + 18.72\,\text{J}$

   $= 18.72\,\text{J}$

   In part c) $Q = -14.0\,\text{J}$.

   Answer

   a) $W \approx 18.7\,\text{J}$. b) $\Delta U \approx 18.7\,\text{J}$, three significant figures. c) $Q = -14.0\,\text{J}$. In the quick compression there is no time for heat to be transferred out, so all the work stays as internal energy of the air and the temperature rises. In the slow compression the same energy is transferred as heat through the walls of the cylinder to the surroundings, so the internal energy and the temperature stay the same.

   Reasonableness

   The average force is

   $\dfrac{18.72\,\text{J}}{0.20\,\text{m}}$$\approx 94\,\text{N}$

   which lies between the initial force $50\,\text{N}$ and the final force $160\,\text{N}$. In the quick compression the internal energy increases more than the work needed in the slow one: $18.7\,\text{J} > 14.0\,\text{J}$. This fits the fact that when the piston is pushed quickly the pump gets hot and the air pressure rises higher, so the compression needs a larger force.

5. Exercise 6-20

   The electrical energy $E_\text{v}$ going to the heating element of a kettle is measured with an energy meter, and the increase $\Delta U$ in the internal energy of the water has been calculated from the rise in temperature. The table gives, as sample data, the measurements at different moments; the heating element transfers energy to the water as heat, and the water is not stirred and no work is done on it. Assume that the warming of the heating element and of the body of the kettle themselves is negligible. According to the data, energy does not seem to be conserved.

   a) Calculate the mean power $P$ of the heating element over the whole measurement.

   b) Calculate how large a quantity of heat $Q_\text{out}$ has been transferred from the water to the surroundings at the moment $120\,\text{s}$, and what percentage of the energy of the heating element it is.

   c) Using conservation of energy, explain where the “missing” energy has gone and why the first law $\Delta U = Q + W$ still holds in the data. Justify with the table that the accumulated loss grows with time.

   The answer box takes the quantity of heat $Q_\text{out}$ of part b) in kilojoules.

   Energy of the kettle's heating element and increase in the internal energy of the water (sample data)

   $t\,(\text{s})$	$E_\text{v}\,(\text{kJ})$	$\Delta U\,(\text{kJ})$
   0	0.0	0.0
   30	60.1	52.0
   60	119.8	103.5
   90	180.3	154.2
   120	239.9	205.1

   kJCheck
   Answer

   $Q_\text{out} = 34.8\,\text{kJ}$

   Show the solution

   Given

   the system is the water; the table values of $E_\text{v}$ and $\Delta U$ at the moments $0$, $30$, $60$, $90$ and $120\,\text{s}$; $W = 0$; the heating element transfers the energy $E_\text{v}$ to the water as heat

   Find

   a) the mean power $P$, b) the quantity of heat $Q_\text{out}$ transferred from the water to the surroundings at the moment $120\,\text{s}$ and its percentage, c) an explanation using conservation of energy

   Principle

   Energy is conserved, so $\Delta U = Q + W$. The water is not stirred and no work is done on it, so $W = 0$. The warming of the heating element and of the body of the kettle themselves is assumed to be negligible, so the net heat transferred to the water is the energy transferred by the heating element minus the heat escaping to the surroundings: $Q = E_\text{v} - Q_\text{out}$. So $\Delta U = E_\text{v} - Q_\text{out}$ and $Q_\text{out} = E_\text{v} - \Delta U$. Power is energy divided by time: $P = \dfrac{E_\text{v}}{t}$.

   Reading the graph

   In the last row $t = 120\,\text{s}$, $E_\text{v} = 239.9\,\text{kJ}$ and $\Delta U = 205.1\,\text{kJ}$. In the first row both are zero.

   Formula

   $P = \dfrac{E_\text{v}}{t}$, $Q_\text{out} = E_\text{v} - \Delta U$ and the share $= \dfrac{Q_\text{out}}{E_\text{v}} \cdot 100\,\%$

   Conversions

   $E_\text{v}$$= 239.9\,\text{kJ}$

   $= 239\,900\,\text{J}$

   so that the power comes out in watts; $1\,\text{kW} = 1000\,\text{W}$.

   Substitution

   a)

   $P$$= \dfrac{239\,900\,\text{J}}{120\,\text{s}}$

   $= 1999\,\text{W}$

   $\approx 2.00\,\text{kW}$

   b)

   $Q_\text{out}$$= 239.9\,\text{kJ} - 205.1\,\text{kJ}$

   $= 34.8\,\text{kJ}$

   and the share is

   $\dfrac{34.8\,\text{kJ}}{239.9\,\text{kJ}} \cdot 100\,\%$

   $= 14.5\,\%$

   Answer

   $Q_\text{out} = 34.8\,\text{kJ}$

   (part b), three significant figures, like the values of the table, that is, about $14.5\,\%$ of the energy of the heating element. Part a): the mean power is $P = 2.00\,\text{kW}$. Part c): The missing energy has not disappeared: it has been transferred as heat through the wall of the kettle to the surrounding air and the table, whose internal energy has increased by the same amount. The first law holds when the net heat $Q = E_\text{v} - Q_\text{out}$ is used for the water. The accumulated loss $E_\text{v} - \Delta U$ grows on every row of the table: $8.1$, $16.3$, $26.1$ and $34.8\,\text{kJ}$, because the warm water gives off heat to the surroundings all the time.

   Reasonableness

   The loss is smaller than the energy of the heating element ($34.8\,\text{kJ} < 239.9\,\text{kJ}$), so the internal energy of the water still increases, as in the table. Balance: $205.1\,\text{kJ} + 34.8\,\text{kJ} = 239.9\,\text{kJ}$. In the measurement the power is $2.00\,\text{kW}$ and the loss is on average about $0.29\,\text{kW}$ all the time, which is about $15\,\%$ of the power, that is, a clearly measurable part.