Some very hot rocks have a temperature of #240 ^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?

1 Answer
yumean1119
Nov 7, 2017

You need the heat of vaporization to answer the question.
About #1135# kg of rocks are needed.

Explanation:

The heat of vaporization for water is #2257# kJ/kg(#=2.257*10^6# J/kg) (http://www.hakko.co.jp/qa/qakit/html/h01060.htm, Japanese).

Then, let's solve the problem with the formula #Q=mcΔT#.
(#Q#: gained heat(J), #m#: mass of the object(kg), #c#: specific heat(J/kg#*#K), #ΔT#: change in temparature(K))

Let #x# (kg) the minimum combined mass of the rocks.
If #x# (kg) of rocks are cooled from #240#℃ to #100#℃, the rocks will release #x*240*(240-100)=3.36*10^4x# (J) of heat.

And, if #16L# (#=16# kg) of water are heated from #70℃# to #100℃# and then vaporized, water will gain #16*4184*(100-70)+16*2.257*10^6=3.812*10^7# (J) of heat. (#4184#(J/kg#*#K) is the specific heat of water. #1# cal=#4.184# J.)

The heat energy must follow the law of energy conservation law.
Therefore, the mass of rocks is
#3.36*10^4x=3.812*10^7#
#x= 1134.53# (kg).

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