How to find specific latent heat from a heating graph

La domanda

La soluzione

1. 01Read the shape of the graph first. For the first minute the ice warms from $-20\ ^\circ\mathrm{C}$ to $0\ ^\circ\mathrm{C}$, then the line is flat for eight minutes while the sample melts, with energy still going in at the same rate.
2. 02Use two formulas: $Q = m c \Delta\theta$ on the sloping part and $Q = m L$ on the flat part. The units check this: $c$ is in $\mathrm{J\,kg^{-1}K^{-1}}$ and $L$ is in $\mathrm{J\,kg^{-1}}$, so a kelvin in a latent heat line means the wrong formula.
3. 03Part (a): use the first minute, the only part where the temperature changes. $Q = m c \Delta\theta = 0.20 \times 2100 \times 20 = 8400\ \mathrm{J}$.
4. 04That energy took 1 minute, which is 60 s, so $P = \frac{Q}{t} = \frac{8400}{60} = 140\ \mathrm{W}$. The rate is constant, so this value holds for the whole graph.
5. 05Do not use $Q = m c \Delta\theta$ on the flat part. There $\Delta\theta = 0$, so the formula gives zero joules for the eight minutes in which most of the energy was transferred.
6. 06Part (b): the flat part runs from 1.0 min to 9.0 min, so $t_{\text{melt}} = 9.0 - 1.0 = 8.0\ \text{min} = 480\ \mathrm{s}$. Because the rate is constant, $Q = P t = 140 \times 480 = 6.72 \times 10^{4}\ \mathrm{J}$.
7. 07Divide by the mass: $L = \frac{Q}{m} = \frac{6.72 \times 10^{4}}{0.20} = 3.4 \times 10^{5}\ \mathrm{J\,kg^{-1}}$. The data booklet value for ice is $3.34 \times 10^{5}\ \mathrm{J\,kg^{-1}}$, which confirms the result.
8. 08Part (c): internal energy is $U = E_{\text{k}} + E_{\text{p}}$, the kinetic energy of the molecules plus the potential energy stored in the bonds between them. The temperature measures only the kinetic part.
9. 09At 0.5 min the sample is on the slope, so $E_{\text{k}}$ is rising. At 5.0 min it is on the flat part, so $E_{\text{k}}$ is constant while $E_{\text{p}}$ rises as the bonds break. The internal energy is greater at 5.0 min, for a reason the thermometer cannot show.

Risposta
The latent heat of fusion of ice is 3.4 × 10⁵ J kg⁻¹ (P = 140 W); internal energy is greater at 5.0 min than at 0.5 min.