An object with a mass of #1 kg#, temperature of #350 ^oC#, and a specific heat of #14 J/(kg*K)# is dropped into a container with #28 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 10, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.042ºC#

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=350-T#

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

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

#C_o=0.014kJkg^-1K^-1#

#1*0.014*(350-T)=28*4.186*T#

#200-T=(28*4.186)/(1*0.014)*T#

#350-T=8372T#

#8373T=350#

#T=350/8373=0.042ºC#

As #T<100ºC#, the water does not evaporate.