Question

Some very hot rocks have a temperature of `210 ^o C` and a specific heat of `150 J/(Kg*K)`. The rocks are bathed in `121 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

30384 kg

Explanation:

121L of water is at 100°C. Mass of water (`m_2`) = Volume of water`xx`Density
`m_2 = 121L xx 1 kgL^-1 = 121 kg`

121kg of water is completely vaporised. For this heat absorbed by water is `m_2L` where L = latent heat of vapourisation of water. It’s value is 2260 kJ/kg

Heat released by rocks = Heat absorbed by water

`m_1 S ΔT = m_2 L`

`m_1 = (m_2 L) / (S ΔT)`

`m_1 = (121kg xx 2260 kJ*kg^-1) / (0.150 kJkg^-1K^-1 * 60K)`

`m_1 = 30384 kg`