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?

Question

1068 views around the world

You can reuse this answer
Creative Commons License

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#