An object with a mass of #16 kg#, temperature of #180 ^oC#, and a specific heat of #5 J/(kg*K)# is dropped into a container with #48 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 21, 2018

Answer:

The water does not evaporate and the change in the tempertaure is #=0.07^@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=180-T#

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

The specific heat of the object is #C_o=0.005kJkg^-1K^-1#

The mass of the object is #m_0=16kg#

The volume of water is #V=48L#

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

The mass of the water is #m_w=rhoV=48kg#

#16*0.005*(180-T)=48*4.186*T#

#180-T=(48*4.186)/(16*0.005)*T#

#180-T=2511.6T#

#2512.6=180#

#T=180/2512.6=0.07^@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.