An object with a mass of #3 kg#, temperature of #331 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #42 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 #=56.3#º

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=331-T#

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

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

#C_o=12kJkg^-1K^-1#

#3*12*(331-T)=42*4.186*T#

#331-T=(42*4.186)/(3*12)*T#

#331-T=4.88T#

#5.88T=331#

#T=331/5.88=56.3ºC#

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