An object with a mass of #3 kg#, temperature of #244 ^oC#, and a specific heat of #37 (KJ)/(kg*K)# is dropped into a container with #16 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Jun 18, 2018

Answer:

Some water will evaporate.

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=244-T#

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

The specific heat of the object is #C_o=37Jkg^-1K^-1#

The mass of the object is #m_0=3kg#

The volume of water is #V=16L#

The density of water is #rho=1kgL^-1#

The mass of the water is #m_w=rhoV=16kg#

#3*37*(244-T)=16*4.186*T#

#244-T=(16*4.186)/(3*37)*T#

#244-T=0.603T#

#1.603T=244#

#T=244/1.603=152.2^@C#

As the final temperature is #T>100^@C#, some water will evaporate.