Some very hot rocks have a temperature of #210 ^o C# and a specific heat of #150 J/(Kg*K)#. The rocks are bathed in #121 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
Aug 9, 2017

30384 kg

Explanation:

121L of water is at 100°C. Mass of water (#m_2#) = Volume of water#xx#Density
#m_2 = 121L xx 1 kgL^-1 = 121 kg#

121kg of water is completely vaporised. For this heat absorbed by water is #m_2L# 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#