An object with a mass of #6 kg#, temperature of #50 ^oC#, and a specific heat of #4 J/(kg*K)# is dropped into a container with #24 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 3, 2018

Answer:

The water will not evaporate and the change in temperature is #=0.012^@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=50-T#

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

The [
The specific heat of the object is #C_o=0.004kJkg^-1K^-1#

The mass of the object is #m_0=6kg#

The volume of water is #V=24L#

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

The mass of the water is #m_w=rhoV=24kg#

#6*0.004*(50-T)=24*4.186*T#

#50-T=(24*4.186)/(6*0.004)*T#

#50-T=4186T#

#4187T=50#

#T=50/4187=0.012^@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 #<100^@C# and the specific heat is low.