An object with a mass of #18 kg#, temperature of #240 ^oC#, and a specific heat of #8 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
Apr 17, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.17ºC#

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=0.008kJkg^-1K^-1#

#18*0.008*(240-T)=48*4.186*T#

#240-T=(48*4.186)/(18*0.008)*T#

#240-T=1395.3T#

#1396.3T=240#

#T=240/1396.3=0.17ºC#

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