An object with a mass of #4 kg#, temperature of #341 ^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
Mar 6, 2017

Answer:

The water does not evaporate and the temperature changes by #73.1ºC#

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=341-T#

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

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

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

#m_0 C_o*(341-T) = m_w* 4.186 *T#

#4*12*(341-T)=42*4.186*T#

#341-T=(42*4.186)/(48)*T#

#341-T=3.66T#

#4.66T=341#

#T=341/4.66=73.1ºC#

The water does not evaporate