An object with a mass of #3 kg#, temperature of #123 ^oC#, and a specific heat of #14 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 4, 2018

Answer:

The water will not evaporate and the change in temperature is #=0.04^@C#

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=123-T#

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

The specific heat of the object is #C_o=0.014kJkg^-1K^-1#

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

The volume of water is #V=32L#

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

The mass of the water is #m_w=rhoV=32kg#

#3*0.014*(123-T)=32*4.186*T#

#123-T=(32*4.186)/(3*0.014)*T#

#123-T=3189.3T#

#3190.3T=123#

#T=123/3190.3=0.04^@C#

As the final temperature is #T<100^@C#, the water will not evaporate. We expect this result as the temperature of the object is not very high and the specific heat is low.