An object with a mass of #24 kg#, temperature of #210 ^oC#, and a specific heat of #8 J/(kg*K)# is dropped into a container with #48 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 · 2017-05-04T13:17:21

The water does not evaporate and the change in temperature is #=0.2ºC#

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=210-T#

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

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

#C_o=0.008kJkg^-1K^-1#

#24*0.008*(210-T)=48*4.186*T#

#210-T=(48*4.186)/(24*0.008)*T#

#210-T=1046.5T#

#1047.5T=210#

#T=210/1047.5=0.2ºC#

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