Some very hot rocks have a temperature of 370 ^o C370oC and a specific heat of 90 J/(Kg*K)90JKgK. The rocks are bathed in 108 L108L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
Nov 18, 2017

The mass of the rocks is =10031kg=10031kg

Explanation:

The heat transferred from the hot rocks to the water, is equal to the heat used to evaporate the water.

For the rocks DeltaT_o=370-100=270^@

Heat of vaporisation of water

H_w=2257kJkg^-1

The specific heat of the rocks is

C_o=0,09kJkg^-1K^-1

The mass of water is m_w=108kg

m_o C_o (DeltaT_o) = m_w H_w

m_o*0.09*270=108*2257

The mass of the rocks is

m_o=(108*2257)/(0,09*270)

=10031kg