An object with a mass of #6 kg#, temperature of #315 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #39 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Jul 25, 2017

Answer:

The water does not evaporate and the change in temperature is #=96.4^@C#

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=315-T#

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

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

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

#6*12*(315-T)=39*4.186*T#

#315-T=(39*4.186)/(6*12)*T#

#315-T=2.267T#

#3.267T=315#

#T=315/3.267=96.4^@C#

As #T<100^@C#, the water does not evaporate