An object with a mass of #4 kg#, temperature of #265 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #39 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Jan 26, 2018

The water will not evaporate and the change in temperature is #=60.2^@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=265-T#

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

The specific heat of water is #C_w=4.186kJkg^-1K^-1#

The specific heat of the object is #C_o=12kJkg^-1K^-1#

The mass of the object is #m_0=4kg#

The mass of the water is #m_w=39kg#

#4*12*(265-T)=39*4.186*T#

#265-T=(39*4.186)/(4*12)*T#

#265-T=3.4T#

#4.4T=265#

#T=265/4.4=60.2^@C#

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