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
Narad T.
Mar 4, 2018

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.

Related questions
See all questions in Specific Heat
Impact of this question
1250 views around the world
You can reuse this answer
Creative Commons License