Some very hot rocks have a temperature of #160 ^o C# and a specific heat of #90 J/(Kg*K)#. The rocks are bathed in #30 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
Jul 29, 2017

The mass of the rocks is #=12539kg#

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=160-100=60^@C#

Heat of vaporisation of water

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0,09kJkg^-1K^-1#

The mass of water is #m_w=30kg#

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.09*60=30*2257#

The mass of the rocks is

#m_o=(30*2257)/(0,09*60)#

#=12539kg#