Quantity of Heat

During fusion, heat only loosens the molecules from their fixed lattice positions; they stay close together and the volume changes very little. During vaporization, the molecules must be pulled completely apart against strong intermolecular attraction, and the large volume increase does extra work against the atmospheric pressure. Much more energy is needed per kilogram, so $L_v \gg L_f$ (for water, $2.26\times10^6$ against $3.36\times10^5\ \text{J kg}^{-1}$).

By the principle of calorimetry, heat lost by the aluminium equals heat gained by the water and the calorimeter. The temperature rise of water and calorimeter is $22.5-15=7.5\ ^\circ\text{C}$, and the aluminium falls by $100-22.5=77.5\ ^\circ\text{C}$.

Heat gained by the calorimeter: $0.3\times400\times7.5=900\ \text{J}$. Heat gained by the water: $0.5\times4200\times7.5=15750\ \text{J}$. Total gained $=16650\ \text{J}$.

Heat lost by the aluminium:

$0.56\times c\times 77.5=43.4\,c$

Equating,

$c=\frac{16650}{43.4}\approx 384\ \text{J kg}^{-1}\text{K}^{-1}$

Triple point: the unique temperature and pressure at which the solid, liquid and vapour phases of a substance coexist in equilibrium. For water it is $273.16\ \text{K}$ at $611\ \text{Pa}$ ($0.01^\circ$C).

Specific heat capacity of a solid by the method of mixture:

The principle is that, with no heat loss, heat lost by the hot body equals heat gained by the cold body and its container.

1. Weigh a dry calorimeter with stirrer: mass $m_c$, specific heat $c_c$. Fill it about one-third with water and weigh again to get the mass of water $m_w$. Note its temperature $\theta_1$.
2. Weigh the solid (mass $m_s$), tie it with thread and heat it in boiling water (or a steam heater) until it reaches a steady temperature $\theta_2$.
3. Quickly transfer the solid into the calorimeter, cover it, stir gently and note the highest steady temperature $\theta$.

Heat lost by solid = heat gained by water and calorimeter:

$m_sc_s(\theta_2-\theta)=(m_wc_w+m_cc_c)(\theta-\theta_1)$ $c_s=\frac{(m_wc_w+m_cc_c)(\theta-\theta_1)}{m_s(\theta_2-\theta)}$

Heat loss is reduced by using a polished calorimeter in a felt-lined jacket, transferring the solid quickly and choosing a small initial temperature rise.

(a) For sweat to evaporate, its molecules need latent heat of vaporisation. This heat is taken from the body (the skin), so the body loses energy and its temperature falls.

(b) Fall in temperature $=37.2-36.7=0.5^\circ$C.

$Q=mc\Delta\theta=60\times4000\times0.5=1.2\times10^{5}\ \text{J}$

(c) All this energy goes into evaporating sweat, $Q=mL$:

$m=\frac{1.2\times10^{5}}{2.4\times10^{6}}=0.05\ \text{kg}=50\ \text{g}$

(a)(i) $Q=mc\Delta\theta=0.20\times4200\times20=1.68\times10^{4}\ \text{J}$

(ii) Average rate of energy loss: $P=\frac{Q}{t}=\frac{16800}{10\times60}=28\ \text{J s}^{-1}$

(b)(i) Energy to be removed to freeze the water at $0^\circ$C: $Q=mL=0.20\times3.3\times10^{5}=6.6\times10^{4}\ \text{J}$ Assuming the same rate of $28\ \text{J s}^{-1}$, $t=\frac{6.6\times10^{4}}{28}\approx2.4\times10^{3}\ \text{s}\approx39\ \text{min}$

(ii) Assumptions: the freezer extracts energy at the same constant rate as before; the water is at a uniform temperature, and does not supercool; the tray absorbs negligible energy; no energy is gained from the surroundings.

The specific latent heat of fusion is the quantity of heat required to change unit mass of a solid into liquid at its melting point without any change in temperature. Its SI unit is $\text{J kg}^{-1}$.

(a) Specific heat capacity is the heat needed to raise the temperature of 1 kg (or 1 g) of a substance by 1 K (or $1^\circ$C); heat is supplied and the temperature changes, $Q=ms\Delta\theta$. Latent heat is the heat needed to change the state of 1 kg of the substance at constant temperature; heat is supplied but the temperature stays constant, $Q=mL$. Specific heat is used within a single state, latent heat is for a change of state (it is used to break or loosen intermolecular bonds, not to raise kinetic energy).

(b) Four stages, with $m=12.0$ g:

• Ice $-10^\circ$C to $0^\circ$C: $Q_1=12\times0.5\times10=60$ cal
• Melting at $0^\circ$C: $Q_2=12\times80=960$ cal
• Water $0^\circ$C to $100^\circ$C: $Q_3=12\times1\times100=1200$ cal
• Boiling at $100^\circ$C: $Q_4=12\times540=6480$ cal

$Q=60+960+1200+6480=8700\ \text{cal}$

Using $1\ \text{cal}=4.2\ \text{J}$: $Q=8700\times4.2\approx 3.65\times10^{4}\ \text{J}$.

So the heat required is $8700$ cal $\approx 36\,540$ J.

(a) It means that $2.26\ \text{MJ}$ ($2.26\times10^{6}\ \text{J}$) of heat is needed to change $1\ \text{kg}$ of water at its boiling point completely into steam at the same temperature, with no change in temperature.

(b) Given $P=1.0\ \text{kW}=1000\ \text{W}$, $m=500\ \text{g}=0.5\ \text{kg}$, $L=2.26\times10^{6}\ \text{J kg}^{-1}$.

$Q=mL=0.5\times2.26\times10^{6}=1.13\times10^{6}\ \text{J}$

$t=\frac{Q}{P}=\frac{1.13\times10^{6}}{1000}=1130\ \text{s}\approx 18.8\ \text{min}$

(c) Not all of the electrical energy goes into the water. Some heat is lost to the surroundings by conduction, convection and radiation, and some is used to heat the kettle itself. More energy must therefore be supplied than $mL$, which takes a little longer.

It means that $2.26\ \text{MJ}$ ($2.26\times10^{6}\ \text{J}$) of heat energy is required to convert $1\ \text{kg}$ of water at its boiling point ($100^\circ\text{C}$) completely into steam at the same temperature, without any change in temperature.

Given $P=1.0\ \text{kW}=1000\ \text{W}$, $m=500\ \text{g}=0.5\ \text{kg}$, $L=2.26\times10^{6}\ \text{J kg}^{-1}$. The water is already boiling, so all the heat goes into vaporisation.

$Q=mL=0.5\times2.26\times10^{6}=1.13\times10^{6}\ \text{J}$

$t=\frac{Q}{P}=\frac{1.13\times10^{6}}{1000}=1130\ \text{s}\approx 18.8\ \text{min}$

The kettle takes about $1130\ \text{s}$ to evaporate all the water.

The calculation assumes that all the electrical energy goes into vaporising the water. In practice some heat is lost to the surroundings (conduction, convection and radiation) and some heats the kettle itself, so the water receives energy more slowly than $1.0\ \text{kW}$. More time is needed to supply the $1.13\times10^{6}\ \text{J}$, so the actual time is a little longer than calculated.

Principle of calorimetry: when two bodies at different temperatures are in thermal contact and no heat is lost to the surroundings, the heat lost by the hot body equals the heat gained by the cold body.

Specific latent heat of fusion of ice by the method of mixture.

Let a calorimeter (with stirrer) of mass $m_c$ and specific heat $c_c$ contain water of mass $m_w$ (specific heat $c_w$) at temperature $\theta_1$. A piece of dry ice of mass $m_i$ at $0^\circ$C is added, and the mixture is stirred until all the ice melts, with final temperature $\theta$.

Heat lost by the calorimeter and water:

$Q_1=(m_wc_w+m_cc_c)(\theta_1-\theta)$

Heat gained by the ice: first to melt it, then to warm the melt water from $0$ to $\theta$:

$Q_2=m_iL_f+m_ic_w\,\theta$

By the principle of calorimetry, $Q_1=Q_2$:

$L_f=\frac{(m_wc_w+m_cc_c)(\theta_1-\theta)}{m_i}-c_w\theta$

All the masses are found by weighing and the temperatures with a thermometer, so $L_f$ can be calculated. Its value is about $3.36\times10^5\ \text{J kg}^{-1}$.