Some very hot rocks have a temperature of #360 ^o C# and a specific heat of #140 J/(Kg*K)#. The rocks are bathed in #144 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

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Answer 1 · 2018-01-12T13:09:50

The mass of the rocks is #=8928.8kg#

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=360-100=260^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.14kJkg^-1K^-1#

The mass of water is #m_w=144kg#

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.14*260=144*2257#

The mass of the rocks is

#m_o=(144*2257)/(0.14*260)#

#=8928.8kg#