An object with a mass of #6 kg#, temperature of #50 ^oC#, and a specific heat of #4 J/(kg*K)# is dropped into a container with #27 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-06-11T07:29:44

The water does not evaporate and the change in temperature is #=0.01ºC#

The heat is transferred from the hot object to the cold water.

Let #T=# final temperature of the object and the water

For the cold water, # Delta T_w=T-0=T#

For the object #DeltaT_o=50-T#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

#C_o=0.004kJkg^-1K^-1#

#6*0.004*(50-T)=27*4.186*T#

#50-T=(27*4.186)/(6*0.004)*T#

#50-T=4709.25T#

#4710.25T=50#

#T=50/4710.25=0.01ºC#

As #T<100ºC#, the water does not evaporate