$ 解：(1)Q_吸=cm\Delta t=4.2×10^3\ \text{J/(kg}\cdot ℃) ×{ 1000 }\ \text{kg}×(100℃-20℃)={ 3.36×10^8 }\ \text{J}$
$ (2)Q_放=\frac {Q_吸}η=\frac {3.36×10^8\ \text{J}}{60\%}=5.6×10^8\ \text{J}$
$ V=\frac {Q_放}q=\frac {{ 5.6 }×10^8\ \text{J}}{{ 3.9×10^7 }\ \text{J/m}^3}={ 14.4 }\ \text{m}^3$
