Some very hot rocks have a temperature of 280 ^o C280oC and a specific heat of 40 J/(Kg*K)40JKgK. The rocks are bathed in 60 L60L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
Aug 11, 2017

The mass of the rocks is =18808kg=18808kg

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=280-100=180^@C

Heat of vaporisation of water

H_w=2257kJkg^-1

The specific heat of the rocks is

C_o=0,04kJkg^-1K^-1

The mass of water is m_w=60kg

m_o C_o (DeltaT_o) = m_w H_w

m_o*0.04*180=60*2257

The mass of the rocks is

m_o=(60*2257)/(0,04*180)

=18808kg