An object with a mass of #12 kg#, temperature of #175 ^oC#, and a specific heat of #22 J/(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
Jun 6, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.31ºC#

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=275-T#

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

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

#C_o=0.022kJkg^-1K^-1#

#12*0.022*(175-T)=35*4.186*T#

#175-T=(35*4.186)/(12*0.022)*T#

#175-T=554.96T#

#555.96T=175#

#T=175/555.96=0.31ºC#

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