Some very hot rocks have a temperature of #360 ^o C# and a specific heat of #240 J/(Kg*K)#. The rocks are bathed in #12 L# of water at #75 ^oC#. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
May 8, 2018

The mass of the rocks is #=68.4kg#

Explanation:

The heat transferred from the hot rocks to the water, is equal to the heat used to raise the temperature to #100^@C# and to evaporate the water.

For the rocks #DeltaT_o=360-75=285^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.240kJkg^-1K^-1#

Volume of water is #V=12L#

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

The mass of water is #m_w=Vrho=12kg#,

The temperature of the water is #T_w=75^@C#

# m_o C_o (DeltaT_o) = m_w H_w +m_C_wDeltaT#

#m_o*0.240*285=12*2257+12*4.186*25#

The mass of the rocks is

#m_o=(12*285+12*4.186*25)/(0.240*285)#

#=68.4kg#