An object with a mass of #3 kg#, temperature of #234 ^oC#, and a specific heat of #37 (KJ)/(kg*K)# is dropped into a container with #15 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 12, 2017

The water evaporates

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=234-T#

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

#C_w=4.186kJkg^-1K^-1#

#C_o=37kJkg^-1K^-1#

#3*37*(234-T)=15*4.186*T#

#234-T=(15*4.186)/(111)*T#

#234-T=0.566T#

#1.566T=234#

#T=234/1.566=149.5ºC#

The water evaporates