Question

Some very hot rocks have a temperature of `480 ^o C` and a specific heat of `120 J/(Kg*K)`. The rocks are bathed in `24 L` of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Answer 1

The mass of the rocks is `=1187.9kg`

Explanation:

The heat transferred from the rocks to the water, is equal to the heat used to evaporate the water.

For the rocks `DeltaT_o=480-100=380º`

Heat of vaporisation of water

`H_w=2257kJkg^-1`

The specific heatof the rocks is

`C_o=0.12kJkg^-1K^-1`

The mass of water is `m_w=24kg`

`m_o C_o (DeltaT_o) = m_w H_w`

`m_o*0.12*380=24*2257`

The mass of the rocks is

`m_o=(24*2257)/(0.12*380)`

`=1187.9kg`