Question

Some very hot rocks have a temperature of `460 ^o C` and a specific heat of `270 ( kJ)/(kg)`. The rocks are bathed in `108 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

`=2.5164kg`

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"=108xx2264.76=244594.08kJ` ......(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 `460^@"C to " 100^@"C"` is given by
`Delta Q_"lost"=m_"rocks"cdot270cdot (460-100)` ......(2)
Equating (1) and (2) and solving for the required quantity
`m_"rocks"cdot270cdot (460-100)=244594.08`
`m_"rocks"=244594.08/(270 xx 360)=2.5164kg`