An object with a mass of #3 kg#, temperature of #185 ^oC#, and a specific heat of #37 (KJ)/(kg*K)# is dropped into a container with #15 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 13, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.33ºC#

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=185-T#

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

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

#C_o=0.037kJkg^-1K^-1#

#3*0.037*(185-T)=15*4.186*T#

#185-T=(15*4.186)/(3*0.037)*T#

#185-T=565.7T#

#566.7T=185#

#T=185/566.7=0.33ºC#

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