An object with a mass of #6 kg#, temperature of #200 ^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 · 2017-04-05T05:44:46

The water does not evaporate and the change in temperature is #=0.04ºC#

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=200-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*(200-T)=27*4.186*T#

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

#200-T=4709.3T#

#4710.3T=200#

#T=200/4710.3=0.04ºC#

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