An object with a mass of #5 kg#, temperature of #123 ^oC#, and a specific heat of #14 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?

1 Answer
Apr 30, 2017

Answer:

The water does not evaporate and the change in temperature is #0.05ºC#

Explanation:

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=123-T#

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

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

#C_o=0.014kJkg^-1K^-1#

#5*0.014*(123-T)=32*4.186*T#

#123-T=(32*4.186)/(5*0.014)*T#

#123-T=2511.6T#

#2512.6T=123#

#T=123/2512.6=0.05ºC#

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