The Absorption of Heat by Solids and Liquids — Question 7

Question 7

A heating curve determines both heat capacity and latent heat

A .250 kg pure sample is heated at constant net input 100 W. It warms through 20.0 K in 40.0 s, remains at its melting point for 150 s, then its liquid warms through 10.0 K in 30.0 s. Neglect work and any capacity outside the sample.

1. Infer solid specific heat, latent heat of fusion and liquid specific heat.

2. If the same experiment used net input 80 W, predict the three interval durations and explain which graph features would remain unchanged.

Question 7 - Solution

A heating curve determines both heat capacity and latent heat

(a) Method and calculation

$$c_s=\frac{100(40)}{.250(20)}=800\,\mathrm{J/(kg\,K)}.$$$$L_f=\frac{100(150)}{.250}=60000\,\mathrm{J/kg},\quad c_l=\frac{100(30)}{.250(10)}=1200\,\mathrm{J/(kg\,K)}.$$

(b) Interpretation and further calculation

For unchanged endpoint states, $P\Delta t=Q$ in each stage.$$\Delta t_s=4000/80=50.0\,\mathrm s,\quad\Delta t_f=15000/80=187.5\,\mathrm s.$$$$\Delta t_l=3000/80=37.5\,\mathrm s.$$The melting temperature and temperature spans remain unchanged; the time axis stretches by $100/80=1.25$.

Physical check

The liquid has a smaller heating slope because its inferred specific heat is larger.