An object with a mass of #4 kg#, temperature of #240 ^oC#, and a specific heat of #15 J/(kg*K)# is dropped into a container with #24L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
May 14, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.14ºC#

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=240-T#

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

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

#C_o=0.015kJkg^-1K^-1#

#4*0.015*(240-T)=24*4.186*T#

#240-T=(24*4.186)/(4*0.015)*T#

#240-T=1674.4T#

#1675.4T=240#

#T=240/1675.4=0.14ºC#

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