An object with a mass of #12 kg#, temperature of #270 ^oC#, and a specific heat of #9 J/(kg*K)# is dropped into a container with #32 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 · 2017-03-23T13:54:06

The water does not evaporate and the change in temperature is #=0.22ºC#

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=270-T#

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

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

#C_o=0.009kJkg^-1K^-1#

#12*0.009*(270-T)=32*4.186*T#

#270-T=(32*4.186)/(12*0.009)*T#

#270-T=1240.3T#

#1241.3T=270#

#T=270/1241.3=0.22ºC#

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