RuukkilabraUpper secondary physics
FY3 · at the end of the module

Apply independently

In these tasks nobody tells you which formula to use. Decide yourself which phenomena and models belong to the situation, solve the task on paper and justify your solution, as in the matriculation examination.

Three problems bring together the phenomena of energy, changes of state and gases from the module. In each one, you decide which phenomena and models apply to the situation.

Solve the task yourself first, on paper or on a computer. You may use a calculator, a data booklet and graphing software. Write out your reasoning, formulas, substitutions and answers with units.

Solar collector 15 points

Made-up example data · The data is calculated from a model and rounded as in a measurement. It is not the result of a real measurement.

Students built a solar collector: a shallow black box with a glass plate on top. The box contains $4.00\,\text{kg}$ of water. The collector is turned towards the Sun. According to the meter, the power of the solar radiation is $800\,\text{W/m}^2$, that is, $800\,\text{W}$ for every square metre perpendicular to the direction of the Sun, and the area of the collector in this direction is $0.50\,\text{m}^2$. The water is stirred all the time, and its temperature is measured every $5\,\text{min}$. The air temperature is $20.0\,{}^\circ\text{C}$.

$t\ (\text{min})$ $T\ ({}^\circ\text{C})$
$0$ $20.0$
$5$ $24.1$
$10$ $28.0$
$15$ $31.6$
$20$ $35.0$
$25$ $38.0$
$30$ $40.9$
$35$ $43.8$
$40$ $46.2$
$45$ $48.5$
$50$ $50.7$
$55$ $52.8$
$60$ $54.7$
  1. a) Draw a $(t, T)$ graph of the data and fit a model to the points. 3 p
  2. b) Determine the power at which the water receives energy from the solar radiation. Justify which part of the graph you use to determine it. 4 p
  3. c) Calculate the efficiency of the collector at the start of the measurement, when the water is still at air temperature. 2 p
  4. d) Why does the rise in temperature slow down? Determine the power of the heat loss at $50\,\text{min}$. 4 p
  5. e) Evaluate two assumptions you used in parts b–d. In which direction does each one affect the results? 2 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

the $(t, T)$ graph of the data

Graph

The horizontal axis shows time $t\ (\text{min})$ and the vertical axis shows the temperature of the water $T\ ({}^\circ\text{C})$. The vertical axis does not have to start from zero. Each row is one separate point.

$10$
$20$
$30$
$40$
$50$
$60$
$20$
$30$
$40$
$50$
$60$
$t\ (\text{min})$
$T\ ({}^\circ\text{C})$
$10$
$0$
$\text{tangent: } 0.85\,\tfrac{\text{K}}{\text{min}}$
$0.43\,\tfrac{\text{K}}{\text{min}}$

Model answer

The points are not on a straight line: at first the temperature rises by about $0.8\,\text{K}$ per minute, but at the end only by about $0.4\,\text{K}$ per minute. So a smooth, levelling-off curve is drawn through the points, not a straight line or a broken line. The auxiliary lines for parts b and d are also drawn in the figure.

b)

Given

mass of the water $m = 4.00\,\text{kg}$, specific heat capacity of water $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and the graph from part a

Find

the power $P$ at which the water receives energy from the solar radiation

Principle

The power the water receives shows in how fast its temperature rises: the quantity of heat is $Q = cm\Delta T$ and the power is $P = Q : \Delta t$. The warmer the water is compared with the air, the more heat is conducted and radiated to the surroundings. At the start of the measurement the water is at air temperature, so then almost all the power received from the Sun heats the water. So the power is determined from the start of the graph, not from the average rise over the whole measurement.

Reading the graph

A tangent is drawn to the curve at $0\,\text{min}$. It passes through the points $(0\,\text{min}$; $20.0\,{}^\circ\text{C})$ and $(10\,\text{min}$; $28.5\,{}^\circ\text{C})$, so the initial rate of warming is

$\dfrac{\Delta T}{\Delta t}$$= \dfrac{8.5\,\text{K}}{10\,\text{min}}$
$= 0.85\,\text{K/min}$

Formula

$P$$= \dfrac{Q}{\Delta t}$
$= cm \cdot \dfrac{\Delta T}{\Delta t}$

Conversions

$0.85\,\text{K/min}$$= 0.85\,\text{K} : 60\,\text{s}$
$= 0.014\,17\,\text{K/s}$

A change in temperature is the same in degrees Celsius and in kelvins.

Substitution

$P$$= 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 4.00\,\text{kg} \cdot 0.014\,17\,\text{K/s}$
$= 237\,\text{W}$

Answer

$P \approx 240\,\text{W}$

Two significant figures, because the slope of the tangent is read from the graph to about two figures.

Reasonableness

The radiation arriving at the collector is $800\,\text{W/m}^2 \cdot 0.50\,\text{m}^2 = 400\,\text{W}$, so the water receives a little over half of it. The rise of $4.1\,\text{K}$ in the first five minutes gives almost the same power:

$4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 4.00\,\text{kg} \cdot 4.1\,\text{K} : 300\,\text{s}$
$\approx 230\,\text{W}$

c)

Given

power received by the water $P \approx 237\,\text{W}$, radiation power per area $800\,\text{W/m}^2$ and area $A = 0.50\,\text{m}^2$

Find

the efficiency $\eta$ of the collector at the start of the measurement, when there are no losses yet

Principle

Efficiency is the ratio of useful power to incoming power. The useful power is the power from part b that the water receives at the start before losses, and the incoming power is the power of the radiation hitting the surface of the collector.

Formula

$P_\text{in} = 800\,\text{W/m}^2 \cdot A$ and $\eta = \dfrac{P}{P_\text{in}}$

Conversions

Not needed, the powers are in watts.

Substitution

$P_\text{in}$$= 800\,\text{W/m}^2 \cdot 0.50\,\text{m}^2$
$= 400\,\text{W}$
$\eta$$= \dfrac{237\,\text{W}}{400\,\text{W}}$
$= 0.59$

Answer

$\eta \approx 0.59$

That is, about $60\,\%$, two significant figures as in the power from part b.

Reasonableness

The efficiency is at most one, because the water cannot receive more power than arrives at the collector. In a real collector it stays below one, because part of the radiation is reflected from the glass and the box, so about $60\,\%$ is believable for a simple collector.

d)

Given

power received from the Sun $P \approx 237\,\text{W}$ and the graph from part a

Find

why the warming slows down and the power of the heat loss $P_\text{loss}$ at $50\,\text{min}$

Principle

Heat is conducted through the walls of the box and the glass at a higher power, the greater the temperature difference between the water and the air ($P = UA\Delta T$), and a warm box also radiates more. As the water warms up, the loss grows, so a smaller and smaller part of the power received from the Sun is left for the water and the temperature rises more and more slowly. The power received from the Sun stays the same, so the loss is the difference between it and the power left for the water.

Reading the graph

The slope at $50\,\text{min}$ is read from the line through the points at $40\,\text{min}$ and $60\,\text{min}$, which is almost parallel to the tangent of the curve:

$\dfrac{54.7\,{}^\circ\text{C} - 46.2\,{}^\circ\text{C}}{20\,\text{min}}$
$= 0.425\,\text{K/min}$

Formula

$P_\text{water}$$= cm \cdot \dfrac{\Delta T}{\Delta t}$

$P_\text{loss} = P - P_\text{water}$

Conversions

$0.425\,\text{K/min}$$= 0.425\,\text{K} : 60\,\text{s}$
$= 0.007\,08\,\text{K/s}$

Substitution

$P_\text{water}$$= 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 4.00\,\text{kg} \cdot 0.007\,08\,\text{K/s}$
$= 119\,\text{W}$
$P_\text{loss}$$= 237\,\text{W} - 119\,\text{W}$
$= 118\,\text{W}$

Answer

$P_\text{loss} \approx 120\,\text{W}$

Two significant figures, because both slopes are read from the graph.

Reasonableness

At $50\,\text{min}$ the water is about $31\,\text{K}$ warmer than the air, and the loss is about half of the power received from the Sun. If it continued at the same rate, the loss would grow equal to the power received when the temperature difference is about $60\,\text{K}$. Then the temperature of the water would stop rising at about $80\,{}^\circ\text{C}$, which is believable for a simple collector.

e)

Find

two assumptions of parts b–d and the direction of their effect

Principle

An assumption is good if its error is small compared with the uncertainty of reading. For each assumption, you think about whether the water would in reality receive more or less energy than was calculated.

Model answer

Any two, with justification:
- The warming of the box and the glass has been left out. Part of the energy received from the Sun also heats the box, so the real power of the collector is slightly higher than calculated.
- No losses were assumed at the start. Losses begin as soon as the water becomes warmer than the air. If the slope is read from a line through the first points and not from the tangent at the start, it is too small, and the power comes out slightly too small.
- The meter reading $800\,\text{W/m}^2$ was assumed to stay the same for the whole hour. If a cloud or the change in the height of the Sun reduced the radiation during the measurement, part of the slowing of the warming is due to that and not to losses, and the loss in part d is too large.

Common mistakes

  • A straight line is fitted through all the points, and the whole-measurement average of $0.58\,\text{K/min}$ is used as the rate of warming. Then the power is too small, because later part of the power goes into losses.
  • The slope is left in the unit $\text{K/min}$, and the power comes out $60$ times too large.
  • In the efficiency, the energy received by the water during the whole hour is compared with the incoming radiation, even though the losses keep changing.
  • In part d, the slowing is explained only with the word "losses" without saying why they grow as the temperature rises.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

The axes show the correct quantities and units, and the scales are evenly spaced. 1 p
Each measurement is marked as a separate point in the right place. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. A smooth, levelling-off curve passes through the points. 1 p

b)4 p

Determines the rate of warming from the start of the measurement and justifies it: when the water is at air temperature, almost no heat is transferred to the surroundings. 1 p
The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The slope at the start from the tangent or the first points, about $0.8\text{–}0.9\,\text{K/min}$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $Q = cm\Delta T$ and $P = Q : t$, that is, $P = cm \cdot \dfrac{\Delta T}{\Delta t}$ 1 p
The answer has the correct unit and a precision that matches the given data. $P \approx 240\,\text{W}$; a value of $220\text{–}250\,\text{W}$ obtained from the starting tangent or from the points of the first $5\text{–}10\,\text{min}$ is accepted. 1 p

c)2 p

Calculates the radiation power arriving at the collector, $800\,\text{W/m}^2 \cdot 0.50\,\text{m}^2 = 400\,\text{W}$, and the efficiency as the ratio of useful power to incoming power. 1 p
The answer has the correct unit and a precision that matches the given data. $\eta \approx 0.59$, that is, about $60\,\%$ (using the result of part b) 1 p

d)4 p

Explains that the greater the temperature difference between the water and the air, the more heat is conducted and radiated to the surroundings. 1 p
Concludes that the power left for the water decreases, so the temperature rises more and more slowly. 1 p
The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The slope at $50\,\text{min}$, about $0.42\,\text{K/min}$. 1 p
Calculates the loss as the difference between the power received from the Sun and the power left for the water, about $120\,\text{W}$. 1 p

e)2 p

Names the first assumption of the measurement or the model and justifies in which direction it changes the result. 1 p
Names the second assumption and justifies in which direction it changes the result. 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Pressure cooker 15 points

The values in the table are measured boiling temperatures of water from data books, rounded.

The lid of a pressure cooker closes tightly. When the water heats up, the pressure in the cooker rises until the valve starts to let steam out. The valve keeps the pressure in the cooker $80\,\text{kPa}$ higher than the outside air pressure of $101\,\text{kPa}$. The air left the cooker with the steam during heating, so there is only water vapour under the lid.

$p\ (\text{kPa})$ $T_\text{b}\ ({}^\circ\text{C})$
$70.2$ $90$
$84.6$ $95$
$101.3$ $100$
$120.9$ $105$
$143.4$ $110$
$169.2$ $115$
$198.7$ $120$
$232.3$ $125$

The table gives the boiling temperature $T_\text{b}$ of water at different pressures $p$.

  1. a) What is the temperature of the water in the cooker when the valve lets steam out? 3 p
  2. b) Explain why water boils at a higher temperature in a pressure cooker than in an open pot, and why food cooks faster in it. 3 p
  3. c) The cooker contains $1.50\,\text{kg}$ of water and also potatoes. When the valve hisses, the power of the hob is set to $300\,\text{W}$. Of this, $80\,\%$ is transferred to the contents of the cooker and the rest to the surroundings. The potatoes have already warmed through to the temperature of the cooker. How much water vaporises in $20$ minutes? 4 p
  4. d) After cooking, someone opens the lid straight away, even though the pressure has not yet dropped. Explain what happens in the cooker. What fraction of the water vaporises? 5 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

the valve keeps a gauge pressure of $80\,\text{kPa}$, outside air pressure $101\,\text{kPa}$ and the boiling temperatures in the table

Find

the temperature $T$ of the water when the valve lets steam out

Principle

Water boils at the temperature that corresponds to the total pressure acting on the surface of the water. When the valve lets steam out, the total pressure in the cooker is the outside air pressure and the gauge pressure of the valve together. The air has already left, so under the lid there is only water vapour and the whole pressure is the pressure of the vapour. The table gives the boiling temperature only at intervals of $5\,{}^\circ\text{C}$, so a value in between is interpolated between the two nearest rows, assuming that the temperature changes evenly between them.

Formula

$p$$= p_0 + \Delta p$
$T$$= T_1 + (T_2 - T_1) \cdot \dfrac{p - p_1}{p_2 - p_1}$

Conversions

Not needed, all pressures are in kilopascals.

Substitution

$p$$= 101\,\text{kPa} + 80\,\text{kPa}$
$= 181\,\text{kPa}$

which is between the rows $169.2\,\text{kPa}$ and $198.7\,\text{kPa}$.

$T$$= 115\,{}^\circ\text{C} + 5\,{}^\circ\text{C} \cdot \dfrac{11.8\,\text{kPa}}{29.5\,\text{kPa}}$
$= 115\,{}^\circ\text{C} + 2.0\,{}^\circ\text{C}$
$= 117\,{}^\circ\text{C}$

Answer

$T \approx 117\,{}^\circ\text{C}$

To the nearest degree: the gauge pressure of the valve is given to two figures, and between the values of the table the curve bends slightly.

Reasonableness

The temperature is between $115\,{}^\circ\text{C}$ and $120\,{}^\circ\text{C}$ of the table and closer to $115\,{}^\circ\text{C}$, just like the pressure lies between theirs. If the gauge pressure $80\,\text{kPa}$ had been read from the table as such, the boiling temperature would have come out below $95\,{}^\circ\text{C}$, which is impossible for a cooker whose pressure is higher than the outside air.

b)

Find

why the boiling temperature is higher and why food cooks faster

Principle

The table shows that the boiling temperature of water increases as pressure increases. The same information is in the vaporisation curve of the phase diagram, but the table is enough for the answer.

Model answer

When it boils, water vaporises as bubbles throughout the liquid. A vapour bubble exists only if the pressure of the vapour inside it is equal to the pressure of the surrounding liquid, which is affected by the pressure at the surface. The vapour pressure increases as the temperature rises, because faster particles leave the liquid more often and hit the wall of the bubble harder. The higher the pressure at the surface, the higher the temperature must be before bubbles can form. In the pressure cooker the pressure is about $1.8$ times the outside air pressure, so the water boils only at about $117\,{}^\circ\text{C}$.
Food cooks when its inner parts get warm enough. Because the water is $17\,{}^\circ\text{C}$ hotter than in an open pot, the temperature difference to the surface of the food is greater. Then heat is conducted into the food faster, and at a higher temperature the changes of cooking also happen faster.

c)

Given

power of the hob $P = 300\,\text{W}$, of which $\eta = 0.80$ is transferred to the contents of the cooker, time $t = 20\,\text{min}$ and specific latent heat of vaporisation of water $r = 2260\,\text{kJ/kg}$

Find

the mass $m$ of the water that vaporises

Principle

The valve keeps the pressure constant, so the contents of the cooker stay at the boiling temperature. The temperature does not rise, and the potatoes are assumed to be already at the temperature of the cooker, so all the energy transferred to the contents is used to vaporise water, and the steam produced escapes through the valve. The mass of the water and the potatoes is not needed: it only determines how long the water lasts.

Formula

$\eta P t = rm$, so $m = \dfrac{\eta P t}{r}$

Conversions

$t$$= 20\,\text{min}$
$= 20 \cdot 60\,\text{s}$
$= 1200\,\text{s}$
$r$$= 2260\,\text{kJ/kg}$
$= 2.26 \cdot 10^6\,\text{J/kg}$

Substitution

$\eta P t$$= 0.80 \cdot 300\,\text{W} \cdot 1200\,\text{s}$
$= 288\,000\,\text{J}$
$m$$= \dfrac{288\,000\,\text{J}}{2.26 \cdot 10^6\,\text{J/kg}}$
$= 0.127\,\text{kg}$

Answer

$m \approx 0.13\,\text{kg}$

Two significant figures as in the fraction $0.80$ and the time $20\,\text{min}$.

Reasonableness

Less than a tenth of the water vaporises, so there is plenty left. The value of $r$ from the table is measured at $100\,{}^\circ\text{C}$, and at $117\,{}^\circ\text{C}$ it is slightly smaller (about $2.21 \cdot 10^6\,\text{J/kg}$). The result $0.13\,\text{kg}$ still does not change.

d)

Given

temperature of the water $117\,{}^\circ\text{C}$, boiling temperature at outside air pressure $100\,{}^\circ\text{C}$, $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ and $r = 2.26 \cdot 10^6\,\text{J/kg}$

Find

what happens and what fraction $\dfrac{m_\text{v}}{m}$ of the water vaporises

Principle

When the lid is opened, the pressure in the cooker drops straight to the outside air pressure, so the boiling temperature drops to $100\,{}^\circ\text{C}$. The water is $17\,{}^\circ\text{C}$ hotter than its boiling temperature, so it boils violently throughout its volume until it has cooled to $100\,{}^\circ\text{C}$. Energy is conserved: the energy needed for the vaporisation comes from the cooling of the water itself. The part that vaporises is small, so the whole mass $m$ can be used as the mass of the cooling water.

Formula

$c m \Delta T = r m_\text{v}$, so

$\dfrac{m_\text{v}}{m}$$= \dfrac{c \Delta T}{r}$

Conversions

$\Delta T$$= 117\,{}^\circ\text{C} - 100\,{}^\circ\text{C}$
$= 17\,\text{K}$

$r = 2.26 \cdot 10^6\,\text{J/kg}$

Substitution

$\dfrac{m_\text{v}}{m}$$= \dfrac{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 17\,\text{K}}{2.26 \cdot 10^6\,\text{J/kg}}$
$= 0.031$

Answer

about $3\,\%$ of the water vaporises in an instant. One significant figure is enough, because the heat of the potatoes and the cooker was left out of the calculation. They also cool down and vaporise more water, so the real fraction is slightly larger.

Reasonableness

After part c there is about $1.4\,\text{kg}$ of water, and about $40\,\text{g}$ of it vaporises. That is about $2.4\,\text{mol}$, and as an ideal gas at $100\,{}^\circ\text{C}$ and a pressure of $101\,\text{kPa}$ it takes up a volume of

$V$$= \dfrac{nRT}{p}$
$\approx 70\,\text{l}$

Tens of litres of hot steam and boiling splashes are released at once, and steam burns worse as it condenses than water at the same temperature. So a pressure cooker may be opened only after the pressure has dropped.

Also accepted

  • In parts c and d the value $r \approx 2.21 \cdot 10^6\,\text{J/kg}$ for $117\,{}^\circ\text{C}$ can be used; the results do not change.
  • If the boiling temperature in part a is read slightly differently (for example $116$–$118\,{}^\circ\text{C}$), part d is marked using that value.

Common mistakes

  • The temperature corresponding to the gauge pressure $80\,\text{kPa}$ is read from the table, even though the boiling temperature depends on the total pressure.
  • In part c the warming of the water is calculated with the formula $Q = cm\Delta T$, even though the temperature does not change, or the whole power of the hob, $300\,\text{W}$, is used.
  • In part d it is assumed that more heat is needed for the boiling, even though the energy comes from the cooling of the water itself.
  • In part b it is only stated that "pressure raises the boiling point", without explaining why.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

Uses the total pressure in the cooker, $101\,\text{kPa} + 80\,\text{kPa} = 181\,\text{kPa}$, and not just the gauge pressure. 1 p
Reads the boiling temperature by interpolating between two values in the table or from a graph drawn from the table. 1 p
The answer has the correct unit and a precision that matches the given data. $T \approx 117\,{}^\circ\text{C}$ 1 p

b)3 p

Links the boiling temperature to pressure: according to the table, at a higher pressure water boils at a higher temperature. 1 p
Explains why pressure matters, for example with the particle model: a vapour bubble stays open only if the pressure of its vapour is equal to the surrounding pressure, and the vapour pressure increases with temperature. 1 p
Explains the faster cooking by the higher temperature: heat moves into the food faster and the cooking proceeds faster. 1 p

c)4 p

Justifies that the pressure and temperature stay constant and the potatoes are already at the temperature of the cooker, so the energy transferred to the contents of the cooker is used to vaporise water. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\eta P t = rm$, rearranged: $m = \dfrac{\eta P t}{r}$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $0.80 \cdot 300\,\text{W} \cdot 1200\,\text{s} = 288\,\text{kJ}$ 1 p
The answer has the correct unit and a precision that matches the given data. $m \approx 0.13\,\text{kg}$ 1 p

d)5 p

The pressure drops to the outside air pressure, so the boiling temperature drops to $100\,{}^\circ\text{C}$. 1 p
The water is hotter than its boiling temperature, so it boils violently until it has cooled to $100\,{}^\circ\text{C}$. 1 p
Conservation of energy: the heat released as the water cools vaporises part of the water, $c m \Delta T = r m_\text{v}$. 1 p
The answer has the correct unit and a precision that matches the given data. $\dfrac{m_\text{v}}{m} \approx 0.03$, that is, about $3\,\%$ 1 p
Names the danger (steam and splashing boiling water burn) or an assumption of the calculation (the heat of the potatoes and the cooker is left out). 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Diving 15 points

A diver is at a depth of $15.0\,\text{m}$ in a lake. There the temperature of the water is $6.0\,{}^\circ\text{C}$, and at the surface it is $20.0\,{}^\circ\text{C}$. The atmospheric pressure is $101\,\text{kPa}$. The diver breathes from a tank through a regulator: the regulator supplies air at the same pressure as in the surrounding water.

  1. a) The volume of an air bubble breathed out by the diver is $4.0\,\text{cm}^3$ at the bottom. What is its volume at the surface? 4 p
  2. b) At a depth of $15.0\,\text{m}$ the diver fills their lungs with $6.0\,\text{l}$ of air. Why is it dangerous to rise to the surface holding your breath? Justify with a calculation. 3 p
  3. c) The volume of the diver's tank is $12.0\,\text{l}$, and its pressure gauge shows a gauge pressure of $200\,\text{bar}$, where $1\,\text{bar} = 100\,\text{kPa}$. At a depth of $15.0\,\text{m}$ the diver uses $20\,\text{l}$ of air per minute at the pressure of that depth. The ascent is started when the gauge shows $50\,\text{bar}$. Assume that the air is equally warm in the tank and in the lungs. How long can the diver stay at this depth? 5 p
  4. d) Evaluate two assumptions of the calculation in part c. Is the real time likely to be shorter or longer than the calculated one? How should the result be used? 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

volume of the bubble at the bottom $V_1 = 4.0\,\text{cm}^3$, depth $h = 15.0\,\text{m}$, temperatures $T_1 = 6.0\,{}^\circ\text{C}$ and $T_2 = 20.0\,{}^\circ\text{C}$, atmospheric pressure $p_0 = 101\,\text{kPa}$, density of water $\rho = 1000\,\text{kg/m}^3$ and $g = 9.81\,\text{m/s}^2$

Find

the volume $V_2$ of the bubble at the surface

Principle

The amount of air in the bubble stays the same. The bubble is small, so it is always at the temperature of the surrounding water, and its pressure is the pressure of the water at that depth: atmospheric pressure and hydrostatic pressure together. From the equation of state of an ideal gas $pV = nRT$ it follows that when $n$ is constant, $\dfrac{pV}{T}$ stays the same. Both pressure and temperature change, so neither can be left out.

Formula

$p_1$$= p_0 + \rho g h$
$\dfrac{p_1 V_1}{T_1}$$= \dfrac{p_2 V_2}{T_2}$

so

$V_2$$= V_1 \cdot \dfrac{p_1}{p_2} \cdot \dfrac{T_2}{T_1}$

where at the surface $p_2 = p_0$

Conversions

$p_0$$= 101\,\text{kPa}$
$= 101\,000\,\text{Pa}$
$T_1$$= (6.0 + 273.15)\,\text{K}$
$= 279.15\,\text{K}$
$T_2$$= (20.0 + 273.15)\,\text{K}$
$= 293.15\,\text{K}$

The volume can stay in cubic centimetres, because the equation has a ratio of volumes.

Substitution

$p_1$$= 101\,000\,\text{Pa} + 1000\,\text{kg/m}^3 \cdot 9.81\,\text{m/s}^2 \cdot 15.0\,\text{m}$
$= 248\,150\,\text{Pa}$
$V_2$$= 4.0\,\text{cm}^3 \cdot \dfrac{248\,150\,\text{Pa}}{101\,000\,\text{Pa}} \cdot \dfrac{293.15\,\text{K}}{279.15\,\text{K}}$
$= 10.3\,\text{cm}^3$

Answer

$V_2 \approx 10\,\text{cm}^3$

Two significant figures as in the volume $4.0\,\text{cm}^3$.

Reasonableness

The pressure falls by a factor of about $2.5$ as the bubble rises, so the volume grows to about $2.5$ times, and the warming increases it by about $5\,\%$ more. The growth of bubbles as they rise can also be seen in diving videos.

b)

Given

lungs full of air, $V_1 = 6.0\,\text{l}$, at pressure $p_1 = 248\,\text{kPa}$ and pressure at the surface $p_2 = 101\,\text{kPa}$

Find

the volume of the air at the surface and why rising while holding your breath is dangerous

Principle

The amount of air stays the same when you hold your breath. The air in the lungs is at body temperature, about $37\,{}^\circ\text{C}$, regardless of depth and the temperature of the water. So the temperature stays constant, and the volume changes according to Boyle's law.

Formula

$p_1 V_1 = p_2 V_2$, so $V_2 = V_1 \cdot \dfrac{p_1}{p_2}$

Conversions

Not needed, the units cancel in the ratio of the pressures.

Substitution

$V_2$$= 6.0\,\text{l} \cdot \dfrac{248\,\text{kPa}}{101\,\text{kPa}}$
$= 14.7\,\text{l}$

Answer

The air would expand to about $15\,\text{l}$ (two significant figures as in $6.0\,\text{l}$). The lungs can hold only the $6.0\,\text{l}$ given in the problem, so the trapped air would stretch them and could tear the lung tissue. While rising, you must never hold your breath: you breathe normally throughout the ascent, so that the expanding air can escape from the lungs.

Reasonableness

The volume grows in the same ratio as the bubble in part a, except for the share due to warming: the temperature of the lungs does not change.

c)

Given

volume of the tank $V_\text{tank} = 12.0\,\text{l}$, gauge readings $200\,\text{bar}$ and $50\,\text{bar}$ of gauge pressure, consumption $q = 20\,\text{l/min}$ at the pressure of the depth $p = 248\,\text{kPa}$

Find

the time $t$ the diver can stay at a depth of $15.0\,\text{m}$

Principle

According to the problem, the temperature of the air is the same in the tank and in the lungs (part d) estimates the effect of warming), so the amount of air is proportional to the product of pressure and volume (Boyle's law). The air used from the tank takes up, at the pressure of the depth, the volume that is breathed. The gauge shows gauge pressure. However, the difference of the total pressures is the same as the difference of the gauge pressures, because the atmospheric pressure cancels.

Formula

$\Delta p_\text{tank} V_\text{tank} = p V$, so

$V$$= \dfrac{\Delta p_\text{tank} V_\text{tank}}{p}$

and $t = \dfrac{V}{q}$

Conversions

$\Delta p_\text{tank}$$= 200\,\text{bar} - 50\,\text{bar}$
$= 150\,\text{bar}$
$= 150 \cdot 100\,\text{kPa}$
$= 15\,000\,\text{kPa}$

The volumes can stay in litres.

Substitution

$V$$= \dfrac{15\,000\,\text{kPa} \cdot 12.0\,\text{l}}{248\,\text{kPa}}$
$= 725\,\text{l}$
$t$$= \dfrac{725\,\text{l}}{20\,\text{l/min}}$
$= 36\,\text{min}$

Answer

$t \approx 36\,\text{min}$

Two significant figures as in the consumption $20\,\text{l/min}$.

Reasonableness

At the surface the same air would last $2.5$ times longer, about $89\,\text{min}$, because each breath takes $2.5$ times the amount of air at depth. A dive of about half an hour matches ordinary recreational dives.

d)

Find

two assumptions, the direction of their effect and the use of the result

Principle

The calculation is a model. For each assumption, you think about whether in reality more or less air is used than in the model.

Model answer

Any two, with justification:
- The consumption is steady. Exertion, cold water and nervousness increase breathing, so the real time is often shorter.
- The whole time is spent at a depth of $15.0\,\text{m}$. During the descent and the ascent the pressure is lower and less air is used, so in this respect the time is longer.
- The temperature stays the same. The air in the tank is at the temperature of the water, but the consumption $20\,\text{l/min}$ is measured in the lungs at body temperature. When it warms up, the same amount of air takes up about $10\,\%$ more space, so less of the tank air is used per minute and the time is about $10\,\%$ longer.

An additional note: at very high pressure air is no longer exactly an ideal gas, and the tank contains slightly less air than the model gives. This is not required in upper secondary school.
The calculated time is a rough estimate, not a guarantee. A dive is planned so that some air is left in reserve, and the ascent is started by the reading of the gauge and not by the clock.

Common mistakes

  • Only the hydrostatic pressure $\rho g h \approx 147\,\text{kPa}$ is used as the pressure, without the atmospheric pressure.
  • In part a, the temperatures are substituted in degrees Celsius, or the change in temperature is ignored without justification.
  • In part b, the temperature of the water is used as the temperature, even though the air in the lungs is at body temperature.
  • In part c, the air in the tank is compared with the pressure at the surface, so the time comes out about $89\,\text{min}$, more than double.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)4 p

Justifies the model: the amount of air in the bubble stays the same, and the bubble is at the temperature of the surrounding water, so $\dfrac{pV}{T}$ stays constant. 1 p
Calculates the pressure at depth: $p = p_0 + \rho g h \approx 248\,\text{kPa}$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $V_2 = V_1 \cdot \dfrac{p_1}{p_2} \cdot \dfrac{T_2}{T_1}$ with temperatures in kelvins 1 p
The answer has the correct unit and a precision that matches the given data. $V_2 \approx 10\,\text{cm}^3$ 1 p

b)3 p

Chooses a constant temperature and justifies it: the air in the lungs is at body temperature regardless of depth. 1 p
The numerical values are substituted with their units, and the calculation is correct. $V_2 = 6.0\,\text{l} \cdot \dfrac{248\,\text{kPa}}{101\,\text{kPa}} \approx 15\,\text{l}$ 1 p
Explains the danger: the lungs cannot expand this much, so the air damages them; while rising you breathe normally and never hold your breath, so that the expanding air can escape from the lungs. 1 p

c)5 p

Uses the given assumption: when the temperature is the same, the amount of air is proportional to the product of pressure and volume (Boyle's law). 1 p
Uses total pressures or justifies why the difference of the gauge pressures, $150\,\text{bar}$, is the same as the difference of the total pressures. 1 p
Calculates the volume of usable air at the pressure of the depth, about $725\,\text{l}$. 1 p
The reasonableness of the result has been assessed. For example: at the surface the same air would last about $2.5$ times longer, because at depth the pressure is about $2.5$ times as high. 1 p
The answer has the correct unit and a precision that matches the given data. $t \approx 36\,\text{min}$ 1 p

d)3 p

The assumption is named, and its effect on the result is justified. The first assumption and the direction of its effect. 1 p
The assumption is named, and its effect on the result is justified. The second assumption and the direction of its effect. 1 p
The conclusion answers the question and is based on the data. The calculated time is only a rough estimate, and some margin is left in the plan. 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).