An object with a mass of #12 kg#, temperature of #125 ^oC#, and a specific heat of #32 (KJ)/(kg*K)# is dropped into a container with #35 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
May 30, 2017

Answer:

The water does not evaporate and the change in temperature is #=90.5ºC#

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=125-T#

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

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

#C_o=32kJkg^-1K^-1#

#12*32*(125-T)=35*4.186*T#

#125-T=(35*4.186)/(12*32)*T#

#125-T=0.38T#

#1.38T=125#

#T=125/1.38=90.5ºC#

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