Question

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?

Answer 1

`9.06` kg.

Explanation:

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.