Heat Energy

(d) As the specific heat capacity of water is the maximum. Water has a high specific heat capacity, so it absorbs a large amount of heat from the engine with only a small rise in its own temperature.

Put a wet cold cloth (cold compress) on the forehead and sponge the body with cool or lukewarm water, changing the cloth often.

Water has a high specific heat capacity and evaporates by taking heat from the body, so the body temperature falls. Keep the patient in a well-ventilated room and give plenty of fluids.

(a) As of figure A.

The upper fixed point is the temperature of steam over water boiling at normal pressure. So the thermometer bulb must be held in the steam just above the water surface, not dipped in the water (and not touching the flask). Water at the bottom can be at a slightly different temperature from the steam because of impurities and the pressure of the liquid, so a bulb dipped in it gives a wrong reading. Set-ups C and D, in which the bulb is in the water, are therefore wrong.

The specific heat capacity of a substance is the amount of heat required to raise the temperature of 1 kg of the substance by 1$^\circ$C (or 1 K).

An alcohol thermometer is used, because alcohol freezes at about $-115^\circ$C, whereas mercury freezes at $-39^\circ$C, so mercury would freeze at the very low temperatures found on high mountains.

The specific heat capacity of water is $4200\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$ (about $4.2\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$).

Given: metal $m_1 = 600$ g $= 0.6$ kg, initial $100^\circ$C; water $m_2 = 1.5$ L $= 1.5$ kg, initial $15^\circ$C; final temperature $20^\circ$C; $s_w = 4200$ J/kg$^\circ$C. Let the specific heat capacity of the metal be $s$.

Heat lost by metal = heat gained by water:

$0.6 \times s \times (100 - 20) = 1.5 \times 4200 \times (20 - 15)$

$48 s = 31500 \Rightarrow s = 656.25\ \text{J/kg}^\circ\text{C}$

The specific heat capacity of the metal is 656.25 J/kg$^\circ$C.

Given: metal $m_1 = 800$ g $= 0.8$ kg, initial $100^\circ$C; water $m_2 = 1.5$ kg, initial $15^\circ$C; final temperature $20^\circ$C; $s_w = 4200$ J/kg$^\circ$C.

Heat lost by metal = heat gained by water:

$0.8 \times s \times (100 - 20) = 1.5 \times 4200 \times (20 - 15)$

$64 s = 31500 \Rightarrow s \approx 492.2\ \text{J/kg}^\circ\text{C}$

The specific heat capacity of the metal is about 492.2 J/kg$^\circ$C.

(i) The graph shows the anomalous expansion of water: water contracts (instead of expanding) when heated from $0^\circ$C to $4^\circ$C, and its density is maximum at $4^\circ$C.

(ii) Heating from $0^\circ$C to $10^\circ$C:

• From $0^\circ$C to $4^\circ$C the density increases and the volume decreases (minimum volume at $4^\circ$C).
• From $4^\circ$C to $10^\circ$C the density decreases and the volume increases.

(iii) In cold countries the surface of a lake cools first. Water at $4^\circ$C is densest and sinks to the bottom, while colder water (below $4^\circ$C) stays on top and freezes. The ice layer is a poor conductor of heat, so the water below does not freeze and stays at about $4^\circ$C. Fish and other aquatic animals therefore survive in the lake throughout winter.

(a) The graph represents the anomalous expansion of water, that is, the variation of the density of water with temperature.

(b) Water has its minimum volume at $4^\circ$C, because its density is maximum there ($1.0000\ \text{g/cm}^3$) and $V = m/\rho$.

(c) When water at $0^\circ$C is warmed to $8^\circ$C, its volume first decreases from $0^\circ$C to $4^\circ$C and then increases from $4^\circ$C to $8^\circ$C.

(d) At $4^\circ$C water has the minimum volume, so the glass is full with the least volume of water. Either heating (above $4^\circ$C) or cooling (below $4^\circ$C) makes the water expand, so its volume becomes larger than the volume of the glass and the water overflows.

(a) Other liquids expand on heating, so their density falls continuously as temperature rises. Water behaves differently: from $0^\circ$C to $4^\circ$C its density increases (it contracts) and above $4^\circ$C it decreases. Its density is maximum ($1000\ \text{kg/m}^3$) at $4^\circ$C.

(b) Because of this, the water at the bottom of a frozen lake stays at $4^\circ$C, so fish and other aquatic life survive in winter.

(c) Volume of a fixed mass of water against temperature (minimum at $4^\circ$C):

Volume
  |
  |*                         *
  |  *                    *
  |     *              *
  |         *  *  *
  |_________________________ Temperature (°C)
  0         4              10

(d) Water cannot be used as a thermometric liquid because it does not expand regularly: between $0^\circ$C and $4^\circ$C it contracts, so the same volume (and the same reading) occurs at two different temperatures. It also freezes at $0^\circ$C and boils at $100^\circ$C, which limits its range, and it is colourless.

(i) The organism (fish) lives at the bottom in $4^\circ$C water because of the anomalous expansion of water (density of water is maximum at $4^\circ$C).

(ii) Water at $0^\circ$C is less dense than water at $4^\circ$C (density is maximum at $4^\circ$C), so the colder water rises and floats at the top, where it freezes into ice.

(iii) Ice floats and forms an insulating layer that stops further heat loss, so the water below does not freeze and remains at $4^\circ$C. Fish and other aquatic organisms can therefore survive in the lake during severe winter.

(iv) Water pipes burst in cold places in winter (also bottles filled with water crack in a freezer), because water expands on freezing and exerts great pressure.

(a) In winter the temperature falls below $0^\circ$C and the water in the pipes freezes. Water expands on freezing (ice occupies about 9% more volume), and the expanding ice exerts great pressure on the pipe walls, so the pipes burst.

(b) Cover the pipes with insulating material (cloth, straw, foam) or bury them deep underground; letting the tap drip slightly or draining the pipes at night also works.

(c) The surface water cools first and freezes into ice, which floats because it is less dense. Ice is a bad conductor of heat, so it protects the water below. Water at $4^\circ$C is the densest and stays at the bottom, so the temperature at the bottom remains about $4^\circ$C. Aquatic animals survive in this liquid water.

1. Water expands on freezing (below $4^\circ$C it expands, and ice has about 9% more volume than water), so frozen water needs more space than the pipe provides.
2. The expanding ice exerts a very large outward pressure on the pipe walls, which cannot bear it and burst.

(a) Substance C cools the fastest, because it has the lowest specific heat capacity (380 J/kg$^\circ$C). For the same heat loss, $\Delta\theta = Q/(ms)$ is largest when $s$ is smallest.

(b) Given: $m = 20$ kg, $s = 900$ J/kg$^\circ$C, $\Delta\theta = 50^\circ$C.

$Q = m s \Delta\theta = 20 \times 900 \times 50 = 9{,}00{,}000\ \text{J}$

Heat to be supplied = 900,000 J (900 kJ).

Equal masses in the sun for equal time absorb equal heat, and $\Delta\theta = Q/(ms)$, so the smaller the rise in temperature, the larger the specific heat capacity.

(i) Liquid Z has the highest specific heat capacity, because it shows the smallest rise in temperature (28$^\circ$C) for the same heat absorbed.

(ii) Liquid Y will cool down faster. It has the lowest specific heat capacity (largest rise, 35$^\circ$C), so for the same heat lost its temperature falls the most.

(ii) B. Point B is at $4^\circ$C, where the curve peaks; water has its maximum density there.

Water has a very high specific heat capacity, so the sea heats up slowly in summer and cools down slowly in winter. The sea therefore keeps the temperature of the nearby air nearly steady (land and sea breezes help), so Mumbai has a moderate climate, with little difference between summer and winter temperatures.

(d) $m = \dfrac{Q}{S \times dt}$. From $Q = m S\, dt$, dividing both sides by $S\, dt$ gives $m = Q/(S\,dt)$.

Specific heat capacity: the amount of heat required to raise the temperature of 1 kg of a substance by $1^\circ$C.

Unit: joule per kilogram per degree Celsius (J/kg$^\circ$C), or J kg$^{-1}$ K$^{-1}$ in SI.

Two effects of heat:

1. Rise in temperature of a body.
2. Expansion (change in volume) of the body. (Other effects: change of state, chemical change.)

Density of water against temperature, $0^\circ$C to $15^\circ$C:

Density (g/cm3)
1.0000 |        *
       |      *   *
       |    *       *
       |  *           *
       | *               *
0.9990 |                      *
       |_________________________ Temperature (°C)
       0     4              10   15

The density rises from $0^\circ$C to a maximum at $4^\circ$C ($1.0000\ \text{g/cm}^3$) and then falls steadily as the temperature goes up to $15^\circ$C.

(a) Given: $m = 1$ kg, $s = 460$ J/kg$^\circ$C, $\Delta\theta = 80 - 20 = 60^\circ$C.

$Q = m s \Delta\theta = 1 \times 460 \times 60 = 27{,}600\ \text{J}$

Heat lost by the iron ball = 27,600 J.

(b) Volume of a fixed mass of water against temperature:

Volume
  |*                          *
  |   *                   *
  |      *            *
  |          *  *  *
  |_______________________________ Temperature (°C)
  0          4                 10

The volume decreases from $0^\circ$C to $4^\circ$C (minimum at $4^\circ$C) and then increases up to $10^\circ$C.

(c) 3.2 J.

Average kinetic energy = total kinetic energy / number of molecules $= \dfrac{3 + 4 + 5 + 2 + 2}{5} = \dfrac{16}{5} = 3.2\ \text{J}$.

Mercury is used as the thermometric liquid in a laboratory thermometer (range about $-10^\circ$C to $110^\circ$C), because it expands uniformly, is opaque and shiny, and does not wet the glass.

From the graph: $s_B = 910$ J/kg$^\circ$C and $s_C = 4200$ J/kg$^\circ$C.

Heat given to B: $Q = m s \Delta\theta = 1 \times 910 \times (50 - 10) = 36{,}400$ J.

The same heat is given to 1 kg of C:

$\Delta\theta = \frac{Q}{m s_C} = \frac{36400}{1 \times 4200} \approx 8.67^\circ\text{C}$

The temperature of C rises by about 8.67$^\circ$C.

(c) Mercury. The masses are equal and they absorb equal heat from the sun, so $\Delta\theta = Q/(ms)$. Mercury has the lowest specific heat capacity (138 J/kg$^\circ$C), so its temperature rises the most.

(i) Substance Q (140 J/kg$^\circ$C) should fill the bulb. A low specific heat capacity means it needs little heat to change its temperature, so it responds quickly to small changes in temperature.

(ii) Substance S (4200 J/kg$^\circ$C) should be used to reduce fever. Its high specific heat capacity lets it absorb a lot of heat from the body with only a small rise in its own temperature.

(d) nature of matter. Specific heat capacity is a property of the material (for example 4200 for water and 460 for iron) and does not depend on mass, volume or temperature.

Given: $m = 700$ g $= 0.7$ kg, $s = 4200$ J/kg$^\circ$C, $\Delta\theta = 40 - 5 = 35^\circ$C.

$Q = m s \Delta\theta = 0.7 \times 4200 \times 35 = 1{,}02{,}900\ \text{J}$

Heat required = 102,900 J.

a. Given: $m = 500$ g $= 0.5$ kg, $s = 4200$ J/kg$^\circ$C, $\Delta\theta = 85 - 15 = 70^\circ$C.

$Q = m s \Delta\theta = 0.5 \times 4200 \times 70 = 1{,}47{,}000\ \text{J}$

Heat required = 147,000 J.

b. Differences between heat and temperature:

Given: $m = 8$ kg, $s = 4200$ J/kg$^\circ$C, $\Delta\theta = 48 - 30 = 18^\circ$C.

$Q = m s \Delta\theta = 8 \times 4200 \times 18 = 6{,}04{,}800\ \text{J}$

Heat needed = 604,800 J.

Given: $m = 0.5$ kg, $s = 4200$ J/kg$^\circ$C, $\Delta\theta = 90 - 30 = 60^\circ$C.

$Q = m s \Delta\theta = 0.5 \times 4200 \times 60 = 1{,}26{,}000\ \text{J}$

Heat required = 126,000 J.

Given: $m = 500$ g $= 0.5$ kg, $Q = 5980$ J, $\Delta\theta = 226 - 100 = 126^\circ$C.

$s = \frac{Q}{m \Delta\theta} = \frac{5980}{0.5 \times 126} = \frac{5980}{63} \approx 94.9\ \text{J/kg}^\circ\text{C}$

The specific heat capacity is about 94.9 J/kg$^\circ$C.

(a) Kerosene has higher temperature than water. Kerosene has a much lower specific heat capacity than water, and $\Delta\theta = Q/(ms)$, so for equal heat and equal mass its temperature rises more.

Aquatic animals (fish, etc.) survive in the water of frozen lakes and ponds in winter, because the water at the bottom stays at $4^\circ$C beneath the floating ice.

(a) The temperature of water in the ocean does not vary much between day and night. Because water has a high specific heat capacity, it heats and cools slowly. (Options b and c are due to anomalous expansion; d is due to latent heat.)

A digital clinical thermometer has a small range (about $32^\circ$C to $42^\circ$C) and boiling water is at $100^\circ$C, so the high temperature would go beyond its range and damage the thermometer's sensor and electronics.

Given: $m_1 = 10$ kg at $90^\circ$C, $m_2 = 20$ kg at $20^\circ$C, $s = 4200$ J/kg$^\circ$C. Let the final temperature be $\theta$.

Heat lost = heat gained:

$10 \times 4200 \times (90 - \theta) = 20 \times 4200 \times (\theta - 20)$

$90 - \theta = 2\theta - 40 \Rightarrow 3\theta = 130 \Rightarrow \theta \approx 43.3^\circ\text{C}$

The final temperature is about 43.3$^\circ$C.

Meaning: Most liquids expand on heating and contract on cooling, but water behaves abnormally between $0^\circ$C and $4^\circ$C: when heated from $0^\circ$C to $4^\circ$C it contracts, and above $4^\circ$C it expands. Water has minimum volume and maximum density at $4^\circ$C. This irregular behaviour is the anomalous expansion of water.

Effect in daily life: Water pipes burst in winter in cold places, since water expands on freezing. (Also, fish survive in frozen lakes.)

(a) thermal expansion. The increase in the size (volume) of an object when it is heated is called thermal expansion.

The graph shows the anomalous expansion of water: the volume decreases from $0^\circ$C to a minimum at $4^\circ$C and then increases on further heating.

Given: $m = 2000$ g $= 2$ kg, $\Delta\theta = 20^\circ$C.

Specific heat capacity $s = \dfrac{Q}{m\Delta\theta}$:

• A: $\dfrac{168000}{2 \times 20} = 4200$ J/kg$^\circ$C
• B: $\dfrac{18400}{2 \times 20} = 460$ J/kg$^\circ$C
• C: $\dfrac{15400}{2 \times 20} = 385$ J/kg$^\circ$C

(a) Liquid A is appropriate for cooling a hot engine, because it has the highest specific heat capacity (4200 J/kg$^\circ$C, the value for water). It absorbs a large amount of heat with a small rise in temperature.

(b) For substance C, $s = 385$ J/kg$^\circ$C, $Q = 18400$ J, $\Delta\theta = 20^\circ$C:

$m = \frac{Q}{s \Delta\theta} = \frac{18400}{385 \times 20} = \frac{18400}{7700} \approx 2.39\ \text{kg}$

The mass of C is about 2.39 kg (about 2390 g).

1 calorie $= 4.2$ joules (more precisely 4.18 J).

Heat energy is the form of energy that flows from a hot body to a cold body because of a temperature difference. It is the total kinetic energy of the molecules of a body.

A liquid thermometer works on the principle of the expansion of a liquid with a rise in temperature: the liquid in the bulb expands when heated and rises along the tube.

A digital thermometer is used instead of a clinical thermometer because:

1. It gives the reading quickly, and the reading is shown directly as numbers, so it is easy to read accurately.
2. It contains no mercury, so there is no risk of glass breaking and mercury poisoning, and it is safer, especially for children.

(a) It means 400 J of heat energy is needed to raise the temperature of 1 kg of substance Y by $1^\circ$C.

(b) From $Q = mc\,\Delta T$, for equal heat and equal mass, $\Delta T \propto \dfrac{1}{c}$. The substance with the largest specific heat capacity heats up least, so Z (910 J/kg$^\circ$C) has the lowest rise in temperature.

(a) It means 910 J of heat energy is needed to raise the temperature of 1 kg of substance C by $1^\circ$C.

(b) From $Q = mc\,\Delta T$, for equal heat and equal mass, $\Delta T \propto \dfrac{1}{c}$. The substance with the smallest specific heat capacity changes temperature most, so A (138 J/kg$^\circ$C) has the highest temperature change.

(a) The water has more heat energy. The heat content depends on mass, specific heat capacity and temperature. Water has five times the mass (50 g against 10 g) and a much larger specific heat capacity (4200 against about 450 J/kg$^\circ$C), which outweighs the nail's higher temperature.

(b) The iron nail has more average kinetic energy of molecules, because it is at the higher temperature ($180^\circ$C against $100^\circ$C), and temperature measures the average kinetic energy of molecules.

For equal masses and equal heat, $Q = mc\,\Delta T$ gives $\Delta T = \dfrac{Q}{mc}$, so the temperature change is inversely proportional to the specific heat capacity.

$\frac{\Delta T_{\text{liquid}}}{\Delta T_{\text{water}}}=\frac{c_{\text{water}}}{c_{\text{liquid}}}=\frac{4200}{140}=30$

So the liquid's temperature rises 30 times as much as water's for the same heat and mass. It heats up and cools down much faster, because its specific heat capacity is much smaller than that of water.

Given: metal $m_1 = 600$ g $= 0.6$ kg, $\theta_1 = 100^\circ$C; water $m_2 = 0.3$ kg, $\theta_2 = 15^\circ$C, $c_w = 4200\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$; final temperature $\theta = 20^\circ$C.

Specific heat capacity of the metal

Heat lost by metal $=$ heat gained by water:

$0.6\times c\times(100-20)=0.3\times 4200\times(20-15)$

$48\,c = 6300\;\Rightarrow\; c = 131.25\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$

Heat to raise the metal's temperature by $20^\circ$C

$Q = mc\,\Delta T = 0.6\times 131.25\times 20 = 1575\ \text{J}$

The specific heat capacity of the metal is about $131\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$, and 1575 J of heat is needed.

The metals have equal mass, equal volume and the same temperature. Heat contained is $Q = m s \Delta\theta$, so with $m$ and $\Delta\theta$ equal, $Q \propto s$. Metal C has the largest specific heat capacity (910 J/kg$^\circ$C), and A the smallest (380).

(a) Metal C has the highest amount of heat energy, since it has the highest specific heat capacity.

(b) Metal C goes deepest into the wax, because it gives out the most heat and melts the most wax.

(c) Metal A cools down the fastest, since it has the lowest specific heat capacity, so a small loss of heat causes a large fall in temperature.

(d) Metal C loses the highest amount of heat while cooling to the temperature of the wax.

(i) Metal A gets cooled faster. For equal masses losing heat at the same rate, the fall in temperature is $\Delta\theta = Q/(ms)$. A has the lowest specific heat capacity (234 J/kg$^\circ$C), so its temperature falls the most.

(ii) Metal C needs the most heat energy. $Q = ms\Delta\theta$, and with equal mass and the same temperature rise, $Q$ is proportional to $s$. C has the highest specific heat capacity (900 J/kg$^\circ$C).

Water has its highest density at 4$^\circ$C (about $1000\ \text{kg/m}^3$ or $1\ \text{g/cm}^3$).

Why water cools engines: Water has a high specific heat capacity (4200 J/kg$^\circ$C), so it absorbs a large amount of heat from the engine with a small rise in its own temperature. It is also cheap and easily available.

Numerical:

Given: $m_1 = 12$ kg at $\theta_1 = 88^\circ$C; $m_2 = 24$ kg at $\theta_2 = 13^\circ$C; $s = 4200$ J/kg$^\circ$C. Let the final temperature be $\theta$.

Heat lost by hot water = heat gained by cold water:

$12 \times 4200 \times (88 - \theta) = 24 \times 4200 \times (\theta - 13)$

$88 - \theta = 2(\theta - 13) \Rightarrow 88 - \theta = 2\theta - 26$

$3\theta = 114 \Rightarrow \theta = 38^\circ\text{C}$

The final temperature of water in bucket C is 38$^\circ$C.