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 #25 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:

Some water evaporates

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)=25*4.186*T#

#341-T=(25*4.186)/(48)*T#

#341-T=2.18T#

#3.18T=341#

#T=341/3.18=107.22ºC#

Some water evaporates