An object with a mass of #1 kg#, temperature of #120 ^oC#, and a specific heat of #32 J/(kg*K)# is dropped into a container with #12 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 18, 2018

Answer:

The water will not evaporate and the change in temperature is #=0.08^@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=120-T#

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

The specific heat of the object is #C_o=0.032kJkg^-1K^-1#

The mass of the object is #m_0=1kg#

The volume of water is #V=12L#

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

The mass of the water is #m_w=rhoV=12kg#

#1*0.032*(120-T)=12*4.186*T#

#120-T=(12*4.186)/(1*0.032)*T#

#120-T=1569.8T#

#1570.8T=120#

#T=120/1570.8=0.08^@C#

As the final temperature is #T<100^@C#, the water will not evaporate.