Some very hot rocks have a temperature of #420 ^o C# and a specific heat of #300 J/(Kg*K)#. The rocks are bathed in #63 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
Nov 28, 2016

1590,4 kg


to boil 1 L of water need 2272 kJ. For 63 L need 143136 kJ. This heat must be given by the rocks through the formula Q = m cp (T2-T1), where cp is the specific heath of the rock and the mass is unknown. Then M = Q/ cp (T2-T1) = 143126 kJ/ ((0,300 kJ/kg K) x (420-100 °C) =1590,4 kg. being a difference is the same to use temperature in K or in °C