# Some very hot rocks have a temperature of 630 ^o C and a specific heat of 270 J/(Kg*K). The rocks are bathed in 45 L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Nov 11, 2017

${M}_{r} = \left(\setminus \frac{{L}_{v}}{c \setminus \Delta T}\right) {M}_{w} = \setminus \frac{2.26 \setminus \times {10}^{6} \frac{J}{K g}}{\left(270 \frac{J}{K g . K}\right) \left(530 K\right)} 45 K g = 710.69 K g$

#### Explanation:

When the rock is dropped in water, its temperature reduces. The water is already at boiling temperature. This means two things -

1. The minimum temperature that the rock can reach is the boiling temperature of water (${100}^{o} C$),
2. All the heat gained by the water is used for phase change from the liquid phase to the vapour phase.

To find the minimum mass of the rock that completely vaporises the water, we have to equate the heat lost by the rock of unknown mass ${M}_{r}$ in reaching the boiling temperature of the water to the heat gained by 45 Litres of boiling water to undergo a phase change and then solve for ${M}_{r}$.

${M}_{r}$ : Minimum mass of rock that completely vaporises the water,
$c$ : Specific heat capacity of the rock
$\setminus \Delta T$ : Temperature change of the rock (in Kelvins) to achieve the boiling temperature of water.

${M}_{w} = \setminus {\rho}_{w} . {V}_{w}$ : Mass of water evaporated,
$\setminus {\rho}_{w}$ : Density of water; $\setminus q \quad {V}_{w}$: Volume of water;
${L}_{v}$ : Latent heat of vaporisation of water.

Heat Lost by Rock: ${Q}_{r o c k} = {M}_{r} c \setminus \Delta T$
Heat gained by water: ${Q}_{w a t e r} = {M}_{w} {L}_{v}$

${Q}_{r o c k} = {Q}_{w a t e r} \setminus q \quad \setminus \rightarrow \setminus q \quad {M}_{r} c \setminus \Delta T = {M}_{w} {L}_{v}$
${M}_{r} = \left(\setminus \frac{{L}_{v}}{c \setminus \Delta T}\right) {M}_{w}$

L_v = 2.26\times10^6 J/(Kg); \qquad c = 270 J/(Kg.K),
$\setminus \Delta T = {T}_{f} - {T}_{i} = {630}^{o} C - {100}^{o} C = {530}^{o} C = 530 K$,

${M}_{w} = \setminus {\rho}_{w} . {V}_{w} = 1 \frac{K g}{L} \setminus \times 45 L = 45 K g$

${M}_{r} = \setminus \frac{2.26 \setminus \times {10}^{6} \frac{J}{K g}}{\left(270 \frac{J}{K g . K}\right) \left(530 K\right)} 45 K g = 710.69 K g$