Question

Some very hot rocks have a temperature of `360 ^o C` and a specific heat of `840 J/(Kg*K)`. The rocks are bathed in `12 L` of water at `45 ^oC`. 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 `=112.8kg`

Explanation:

The heat transferred from the hot rocks to the water, is equal to the heat used to heat the water from `45^@C` to `100^@C` and to evaporate the water.

For the rocks `DeltaT_o=360-45=315^@`

Heat of vaporisation of water is

`H_w=2257kJkg^-1`

The specific heat of the rocks is

`C_o=0.840kJkg^-1K^-1`

Volume of water is `V=12L`

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

The mass of water is `m_w=Vrho=12kg`,

`m_o C_o (DeltaT_o) = m_wC_w(100-45)+ m_w H_w`

`m_o*0.840*315=(12*4.186*55)+(12*2257)`

The mass of the rocks is

`m_o=(12*4.186*55+12*2257)/(0.840*315)`

`=112.8kg`