Some very hot rocks have a temperature of 210 ^o C210oC and a specific heat of 150 J/(Kg*K)150JKgK. The rocks are bathed in 121 L121L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

1 Answer
Aug 9, 2017

30384 kg

Explanation:

121L of water is at 100°C. Mass of water (m_2m2) = Volume of waterxx×Density
m_2 = 121L xx 1 kgL^-1 = 121 kgm2=121L×1kgL1=121kg

121kg of water is completely vaporised. For this heat absorbed by water is m_2Lm2L where L = latent heat of vapourisation of water. It’s value is 2260 kJ/kg

Heat released by rocks = Heat absorbed by water

m_1 S ΔT = m_2 L

m_1 = (m_2 L) / (S ΔT)

m_1 = (121kg xx 2260 kJ*kg^-1) / (0.150 kJkg^-1K^-1 * 60K)

m_1 = 30384 kg