Some very hot rocks have a temperature of #480 ^o C# and a specific heat of #210 J/(Kg*K)#. The rocks are bathed in #18 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 · 2017-11-22T06:12:46

The mass of the rocks #=509.1kg#

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=480-100=380^@#

Heat of vaporisation of water

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.21kJkg^-1K^-1#

The mass of water is #m_w=18kg#

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.21*380=18*2257#

The mass of the rocks is

#m_o=(18*2257)/(0.21*380)#

#=509.1kg#