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?

1 Answer
Mar 22, 2016

=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