Some very hot rocks have a temperature of #560 ^o C# and a specific heat of #120 J/(Kg*K)#. The rocks are bathed in #24 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 · 2018-04-10T13:11:12

The mass of the rocks is #=981.3kg#

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=560-100=460^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.120kJkg^-1K^-1#

Volume of water is #V=24L#

Density of water is #rho=1kgL^-1#

The mass of water is #m_w=Vrho=24kg#,

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.120*460=24*2257#

The mass of the rocks is

#m_o=(24*2257)/(0.120*460)#

#=981.3kg#