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
A08
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#

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