# 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?

Jun 19, 2018

The mass of the rocks is $= 112.8 k g$

#### Explanation:

The heat transferred from the hot rocks to the water, is equal to the heat used to heat the water from ${45}^{\circ} C$ to ${100}^{\circ} C$ and to evaporate the water.

For the rocks $\Delta {T}_{o} = 360 - 45 = {315}^{\circ}$

Heat of vaporisation of water is

${H}_{w} = 2257 k J k {g}^{-} 1$

The specific heat of the rocks is

${C}_{o} = 0.840 k J k {g}^{-} 1 {K}^{-} 1$

Volume of water is $V = 12 L$

Density of water is $\rho = 1 k g {L}^{-} 1$

The mass of water is ${m}_{w} = V \rho = 12 k g$,

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(100 - 45\right) + {m}_{w} {H}_{w}$

${m}_{o} \cdot 0.840 \cdot 315 = \left(12 \cdot 4.186 \cdot 55\right) + \left(12 \cdot 2257\right)$

The mass of the rocks is

${m}_{o} = \frac{12 \cdot 4.186 \cdot 55 + 12 \cdot 2257}{0.840 \cdot 315}$

$= 112.8 k g$