An object with a mass of #3 kg#, temperature of #235 ^oC#, and a specific heat of #24 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 3, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.13ºC#

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=235-T#

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

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

#C_o=0.024kJkg^-1K^-1#

#3*0.024*(235-T)=32*4.186*T#

#235-T=(32*4.186)/(3*0.024)*T#

#235-T=1860.4T#

#1861.4T=235#

#T=235/1861.4=0.13ºC#

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