Some very hot rocks have a temperature of #280 ^o C# and a specific heat of #40 J/(Kg*K)#. The rocks are bathed in #60 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 · 2017-08-11T07:13:45

The mass of the rocks is #=18808kg#

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#