# 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?

Nov 7, 2017

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($= 2.257 \cdot {10}^{6}$ 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$\cdot$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 $x \cdot 240 \cdot \left(240 - 100\right) = 3.36 \cdot {10}^{4} x$ (J) of heat.

And, if $16 L$ ($= 16$ kg) of water are heated from 70℃ to 100℃ and then vaporized, water will gain $16 \cdot 4184 \cdot \left(100 - 70\right) + 16 \cdot 2.257 \cdot {10}^{6} = 3.812 \cdot {10}^{7}$ (J) of heat. ($4184$(J/kg$\cdot$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
$3.36 \cdot {10}^{4} x = 3.812 \cdot {10}^{7}$
$x = 1134.53$ (kg).