An object with a mass of #4 kg#, temperature of #284 ^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
Feb 22, 2017

Answer:

Water will not evaporate, the final temperature is #=89.3#ºC

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by 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=284-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*(270-T) = m_w* 4.186 *T#

#4*12*(284-T)=25*4.186*T#

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

#284-T=2.18T#

#3.18T=284#

#T=284/3.18=89.3ºC#