The Absorption of Heat by Solids and Liquids — Question 5

Question 5

All ice melts before the mixture warms

An insulated vessel with negligible capacity contains .500 kg water at $40.0^\circ$C. Add .0500 kg ice at $0^\circ$C. Use $c_w=4200\,\mathrm{J/(kg\,K)}$ and $L_f=334000\,\mathrm{J/kg}$.

1. Check that all ice melts, then calculate final temperature.

2. Find the greatest ice mass at $0^\circ$C that could be added while keeping the final temperature at or above $20.0^\circ$C.

Question 5 - Solution

All ice melts before the mixture warms

(a) Method and calculation

$$Q_{w\to0}=.500(4200)(40)=84000\,\mathrm J>.0500(334000)=16700\,\mathrm J.$$$$.500(4200)(40-T_f)=16700+.0500(4200)T_f.$$$$T_f=\frac{84000-16700}{2310}=29.1342^\circ\mathrm C.$$

(b) Interpretation and further calculation

At the limiting final temperature,$$.500(4200)(40-20)=m_i[L_f+c_w(20)].$$$$m_{i,\max}=\frac{42000}{334000+84000}=.100478\,\mathrm{kg}.$$Larger ice mass absorbs too much energy to retain the required final temperature.

Physical check

The melted ice becomes additional water and must also warm to the final temperature.