An object with a mass of #7 kg#, temperature of #163 ^oC#, and a specific heat of #17 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
Mar 16, 2017

Answer:

The water does not evaporate and the temperature change by #0.15ºC#

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=163-T#

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

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

#C_o=0.017kJkg^-1K^-1#

#7*0.017*(163-T)=32*4.186*T#

#163-T=(32*4.186)/(0.119)*T#

#163-T=1125.6T#

#1126.6T=163#

#T=163/1126.6=0.15ºC#

The water does not evaporate