Question

Some very hot rocks have a temperature of `240 ^o C` and a specific heat of `240 J/(Kg*K)`. The rocks are bathed in `16 L` of water at `70 ^oC`. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Answer 1

You need the heat of vaporization to answer the question.
About `1135` kg of rocks are needed.

Explanation:

The heat of vaporization for water is `2257` kJ/kg(`=2.257*10^6` J/kg) (http://www.hakko.co.jp/qa/qakit/html/h01060.htm, Japanese).

Then, let's solve the problem with the formula `Q=mcΔT`.
(`Q`: gained heat(J), `m`: mass of the object(kg), `c`: specific heat(J/kg`*`K), `ΔT`: change in temparature(K))

Let `x` (kg) the minimum combined mass of the rocks.
If `x` (kg) of rocks are cooled from `240`℃ to `100`℃, the rocks will release `x*240*(240-100)=3.36*10^4x` (J) of heat.

And, if `16L` (`=16` kg) of water are heated from `70℃` to `100℃` and then vaporized, water will gain `16*4184*(100-70)+16*2.257*10^6=3.812*10^7` (J) of heat. (`4184`(J/kg`*`K) is the specific heat of water. `1` cal=`4.184` J.)

The heat energy must follow the law of energy conservation law.
Therefore, the mass of rocks is
`3.36*10^4x=3.812*10^7`
`x= 1134.53` (kg).