Some very hot rocks have a temperature of #390 ^o C# and a specific heat of #180 J/(Kg*K)#. The rocks are bathed in #9 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Question

1440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T12:34:00

The mass of the rocks #=389.1kg#

The heat transferred from the hot object, is equal to the heat used to evaporate the water.

For the object #DeltaT_o=290º#

# m_o C_o (DeltaT_o) = m_w H_w #

Heat of vaporisation of water

#H_w=2257kJkg^-1#

#C_o=0.18kJkg^-1K^-1#

#m_0 C_o*290 = m_w* 2257 #

#m_o*0.18*290=9*2257#

#m_o=(9*2257)/(0.18*290)#

#=389.1kg#