An object with a mass of #4 kg#, temperature of #261 ^oC#, and a specific heat of #8 (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
Apr 11, 2017

Answer:

The water does not evaporate and thechange in temperature is #=0.051ºC#

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=261-T#

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

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

#C_o=0.008kJkg^-1K^-1#

#4*0.008*(261-T)=39*4.186*T#

#261-T=(39*4.186)/(4*0.008)*T#

#261-T=5101.7T#

#5102.7T=261#

#T=350/8373=0.051ºC#

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