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?

1 Answer
A08
Mar 27, 2016

#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.

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