An object with a mass of #2 kg#, temperature of #311 ^oC#, and a specific heat of #18 (KJ)/(kg*K)# is dropped into a container with #37 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Sep 16, 2017

Answer:

The water does not evaporate and the change in temperature is #=58.7^@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=311-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=18kJkg^-1K^-1#

The mass of the objest is #m_0=2kg#

The mass of the water is #m_w=37kg#

#2*18*(311-T)=37*4.186*T#

#311-T=(37*4.186)/(2*18)*T#

#311-T=4.30T#

#5.30T=311#

#T=311/5.30=58.7^@C#

As #T<100^@C#, the water does not evaporate