# Some very hot rocks have a temperature of 540 ^o C and a specific heat of 240 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?

Mar 27, 2016

$965.1 k g$ rounded to one decimal place.

#### Explanation:

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by
$\Delta Q = m s t$, or $\Delta Q = m L$
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;
$L$ is the latent heat for the change of state and
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

In the given problem heat is lost by rocks and gained by boiling water which vaporizes.

Heat gained by boiling water to change into vapours at the same temperature i.e., ${100}^{\circ} \text{C}$ is given by $\Delta {Q}_{\text{gained}} = m L$
$\Delta {Q}_{\text{gained}} = 45 \times 2264.76 = 101914.2 k J$ ......(1)
Latent heat of vaporization of water is $2264.76 k J / k g$ and mass of 1 liter of water is $1 k g$.

Now heat lost by rocks to cool down from ${540}^{\circ} \text{C to " 100^@"C}$ is given by
$\Delta {Q}_{\text{lost"=m_"rocks}} \cdot 240 \times {10}^{-} 3 \cdot \left(540 - 100\right)$ ......(2)
Equating (1) and (2) and solving for the required quantity
${m}_{\text{rocks}} \cdot 0.240 \cdot 440 = 101914.2$
${m}_{\text{rocks}} = \frac{101914.2}{0.240 \times 440} = 965.1 k g$ rounded to one decimal place.