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?

1 Answer
Jan 4, 2017

1692.7 kg


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 (J/(cm^3))#.

One liter is #1000 cm^3#, so 72L is #72000 cm^3#.

The total heat for vaporization is thus:
#72000 cm^3 * 2257 (J/(cm^3))# = 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:

#(1.625 x 10^8 J)/(96000) #= 1692.7 kg