Some very hot rocks have a temperature of #360 ^o C# and a specific heat of #125 J/(Kg*K)#. The rocks are bathed in #144 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
Apr 7, 2016

Answer:

Mass of rock required = 1,480 kg

Explanation:

This problem will be solved by first calculating the mass of water involved. Then use the latent heat equation to determine how much energy was required to vaporise the water. And finally use the specific heat equation to determine the mass of rock required to deliver that amount of energy over the implied temperature change in the rock.

1. The mass of water.
Density of water is #1000 kg.m^(-3)#
Volume of water in SI units:
Conversion factor: #1L = 0.001 m^3#
#V = 144 × 10^(-3) = 0.144 m^3#

#ρ = m/V ⇒ m = ρV = 1000 × 0.144 = 144 kg#

2. Determine energy required to vaporise the water.
Latent heat of vaporisation of water (#L_v#) is #334 kJ.kg^(-1)#

# ∆Q = mL_v = 144 × 334 = 48,096 kJ#

3. Determine the minimum mass of rock required.
At this point we should note that the maximum possible temperature change in the rocks is from 360ºC to 100ºC as the water is at its boiling point (i.e. it is at 100ºC).
[If more rock mass was available we could achieve the same energy transfer with a smaller temperature change, but we are asked to find the minimum mass required.]

# Δ Q = mc∆θ #
#⇒ m = (∆Q) / (c∆θ) = (48,096 × 10^3) / (125 × (360 – 100)) = 1,480 kg#