Question

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?

Answer 1

`965.1kg` 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
`DeltaQ=mst`, or `DeltaQ=mL`
where `m,s 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_"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^@"C"` is given by `DeltaQ_"gained"=mL`
`DeltaQ_"gained"=45xx2264.76=101914.2kJ` ......(1)
Latent heat of vaporization of water is `2264.76 kJ//kg` and mass of 1 liter of water is `1kg`.

Now heat lost by rocks to cool down from `540^@"C to " 100^@"C"` is given by
`Delta Q_"lost"=m_"rocks"cdot240xx10^-3cdot (540-100)` ......(2)
Equating (1) and (2) and solving for the required quantity
`m_"rocks"cdot0.240cdot 440=101914.2`
`m_"rocks"=101914.2/(0.240 xx 440)=965.1kg` rounded to one decimal place.