The Absorption of Heat by Solids and Liquids — Question 4

Question 4

Partial ice melting requires a phase check before solving for temperature

An insulated cup of negligible heat capacity contains .200 kg water at $20.0^\circ$C. Add .100 kg ice at $-10.0^\circ$C. Use $c_w=4200$, $c_i=2100$ J/(kg K) and $L_f=334000$ J/kg. Equilibrium melting occurs at $0^\circ$C.

1. Determine final temperature and the masses of remaining ice and liquid.

2. Find the minimum initial water temperature that would just melt all the ice, with final temperature $0^\circ$C.

Question 4 - Solution

Partial ice melting requires a phase check before solving for temperature

(a) Method and calculation

$$Q_{w\to0}=.200(4200)(20)=16800\,\mathrm J,\quad Q_{i\to0}=.100(2100)(10)=2100\,\mathrm J.$$$$m_{\mathrm{melt}}=\frac{16800-2100}{334000}=.0440120\,\mathrm{kg}<.100\,\mathrm{kg}.$$Both phases remain at $0^\circ$C: $m_i=.0559880$ kg and $m_w=.2440120$ kg.

(b) Interpretation and further calculation

$$.200(4200)T_{w,\min}=2100+.100(334000).$$$$T_{w,\min}=\frac{35500}{840}=42.2619^\circ\mathrm C.$$A single-phase mixing equation would be invalid for the original run because the available heat cannot finish melting.

Physical check

Check the phase-energy budget first; a negative temperature from an all-liquid assumption would signal the wrong branch.