Question

Some very hot rocks have a temperature of `280 ^o C` and a specific heat of `40 J/(Kg*K)`. The rocks are bathed in `60 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 `=18808kg`

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=280-100=180^@C`

Heat of vaporisation of water

`H_w=2257kJkg^-1`

The specific heat of the rocks is

`C_o=0,04kJkg^-1K^-1`

The mass of water is `m_w=60kg`

`m_o C_o (DeltaT_o) = m_w H_w`

`m_o*0.04*180=60*2257`

The mass of the rocks is

`m_o=(60*2257)/(0,04*180)`

`=18808kg`