Question

Some very hot rocks have a temperature of `320 ^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

About 720 kg

Explanation:

Let us first calculate how much heat would be required to completely vaporize the water

So
`q= mcDeltat+mL_v`
`(16 kg)(4184J/(kgC))(100°C-70 °C)+ (2,260,000 J/"kg")(16 kg)=`
`2,008,320 J+36,160,000 J= 38,168, 320 J`

So now for the rocks, the heat lost should be:
`q=-mcDeltat`
`q= m*-(240J/(kgK))(373 K-593 K)`
`38168320 J= 52800J/(kg)*m`
`m=720 kg`

Assumption made was: There was no more temperature rise after the water was completely vaporized, so that the final temperature of the system was 100 °C or 373 K