Some very hot rocks have a temperature of #260 ^o C# and a specific heat of #90 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 · 2018-02-07T14:43:25

The mass of the rocks is #=22570kg#

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=260-100=160^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.090kJkg^-1K^-1#

Volume of water is #V=144L#

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

The mass of water is #m_w=Vrho=144kg#,

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.090*160=144*2257#

The mass of the rocks is

#m_o=(144*2257)/(0.090*160)#

#=22570kg#