An object with a mass of #3 kg#, temperature of #170 ^oC#, and a specific heat of #22 J/(kg*K)# is dropped into a container with #18 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Jun 27, 2017

Answer:

The water does not evaporate and the change in temperature is #=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=170-T#

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

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

#C_o=0.022kJkg^-1K^-1#

#3*0.022*(170-T)=18*4.186*T#

#170-T=(18*4.186)/(3*0.022)*T#

#170-T=1141.6T#

#1142.6T=170#

#T=170/1142.6=0.15ºC#

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