An object with a mass of #24 kg#, temperature of #150 ^oC#, and a specific heat of #7 J/(kg*K)# is dropped into a container with #48 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Nov 22, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.13^@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=150-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=0.007kJkg^-1K^-1#

The mass of the object is #m_0=24kg#

The mass of the water is #m_w=48kg#

#24*0.007*(150-T)=48*4.186*T#

#150-T=(48*4.186)/(24*0.007)*T#

#150-T=1196T#

#1197T=150#

#T=150/1197=0.13^@C#

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