An object with a mass of #2 kg#, temperature of #244 ^oC#, and a specific heat of #33 (KJ)/(kg*K)# is dropped into a container with #16 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1149 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T18:54:44

The water will evaporate.

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=244-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=33kJkg^-1K^-1#

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

The mass of the water is #m_w=16kg#

#2*33*(244-T)=16*4.186*T#

#244-T=(16*4.186)/(2*33)*T#

#244-T=1.015T#

#2.015T=244#

#T=244/2.015=121.1^@C#

As #T>100^@C#, the water will evaporate