Question

Some very hot rocks have a temperature of `190 ^o C` and a specific heat of `90 J/(Kg*K)`. The rocks are bathed in `270 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 `=75233.3kg`

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=190-100=90^@`

Heat of vaporisation of water is

`H_w=2257kJkg^-1`

The specific heat of the rocks is

`C_o=0.090kJkg^-1K^-1`

Volume of water is `V=270L`

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

The mass of water is `m_w=Vrho=270kg`,

`m_o C_o (DeltaT_o) = m_w H_w`

`m_o*0.090*90=270*2257`

The mass of the rocks is

`m_o=(270*2257)/(0.090*90)`

`=75233.3kg`