# Some very hot rocks have a temperature of 420 ^o C and a specific heat of 300 J/(Kg*K). The rocks are bathed in 72 L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Jan 4, 2017

1692.7 kg

#### Explanation:

The heat energy from the rocks must at a minimum be sufficient to provide the heat of vaporization of water for the specified volume. It is already at the boiling point, so ONLY the energy for the phase change is required. The heat of vaporization for water is 2257 J/g, or $2257 \left(\frac{J}{c {m}^{3}}\right)$.

One liter is $1000 c {m}^{3}$, so 72L is $72000 c {m}^{3}$.

The total heat for vaporization is thus:
$72000 c {m}^{3} \cdot 2257 \left(\frac{J}{c {m}^{3}}\right)$ = 1.625 x 10^8 J

For a “minimum mass” the rocks will have equilibrated at 100’C with the water. Therefore the heat loss by the rocks is (420 – 100) * 300 = 96000 J/kg. (the delta-T is the same whether in ‘C or ‘K) . This is divided into the known heat used in vaporization to obtain the mass:

$\frac{1.625 x {10}^{8} J}{96000}$= 1692.7 kg