Some very hot rocks have a temperature of #220 ^o C# and a specific heat of #160 J/(Kg*K)#. The rocks are bathed in #50 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
Mar 5, 2018

The mass of the rocks is #=5877.6kg#

Explanation:

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=220-100=120^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.16kJkg^-1K^-1#

Volume of water is #V=50L#

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

The mass of water is #m_w=Vrho=50kg#,

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.16*120=50*2257#

The mass of the rocks is

#m_o=(50*2257)/(0.16*120)#

#=5877.6kg#