An object with a mass of #3 kg#, temperature of #155 ^oC#, and a specific heat of #17 (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
Nov 30, 2017

Answer:

The water does not evaporate and the change in temperature is #=69.5^@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=155-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=17kJkg^-1K^-1#

The mass of the object is #m_0=3kg#

The mass of the water is #m_w=15kg#

#3*17*(155-T)=15*4.186*T#

#155-T=(15*4.186)/(3*17)*T#

#155-T=1.231T#

#2.231T=155#

#T=155/2.231=69.5^@C#

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