Some very hot rocks have a temperature of #240 ^o C# and a specific heat of #150 J/(Kg*K)#. The rocks are bathed in #84 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

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Answer 1 · 2017-01-25T07:25:11

#9.06# kg.

Heat lost by rocks and heat gained by water must be equal.

Here we had #84L# of water, i.e. #84# kg. of boiling water.

As each kg. of boiling water requires #2264.76# KJ of heat for conversion to steam

Water has gained #84xx2264.76# KJ of heat i.e. #190239.84# KJ.

Fall in temperature #t_2-t_1# of hot rocks is #240^@-100^@=140^@#

and as Heat lost = #190239.84=mxx150xx(t_2-t_1)#

or #m=190239.84/(150xx140)=9.05904~~9.06# kg.