An object with a mass of #8 kg#, temperature of #210 ^oC#, and a specific heat of #21 J/(kg*K)# is dropped into a container with #36 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 2, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.23ºC#

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=210-T#

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

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

#C_o=0.021kJkg^-1K^-1#

#8*0.021*(210-T)=36*4.186*T#

#210-T=(36*4.186)/(8*0.021)*T#

#210-T=897T#

#898T=210#

#T=210/898=0.23ºC#

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