# Some very hot rocks have a temperature of 360 ^o C and a specific heat of 240 J/(Kg*K). The rocks are bathed in 12 L of water at 75 ^oC. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

May 8, 2018

The mass of the rocks is $= 68.4 k g$

#### Explanation:

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

For the rocks $\Delta {T}_{o} = 360 - 75 = {285}^{\circ}$

Heat of vaporisation of water is

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

The specific heat of the rocks is

${C}_{o} = 0.240 k J k {g}^{-} 1 {K}^{-} 1$

Volume of water is $V = 12 L$

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

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

The temperature of the water is ${T}_{w} = {75}^{\circ} C$

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

${m}_{o} \cdot 0.240 \cdot 285 = 12 \cdot 2257 + 12 \cdot 4.186 \cdot 25$

The mass of the rocks is

${m}_{o} = \frac{12 \cdot 285 + 12 \cdot 4.186 \cdot 25}{0.240 \cdot 285}$

$= 68.4 k g$