An object with a mass of #1 kg#, temperature of #170 ^oC#, and a specific heat of #32 J/(kg*K)# is dropped into a container with #8 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 25, 2017

Answer:

The water does not evaporate and the change in temperature is #0.16ºC#

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=0.032kJkg^-1K^-1#

#1*0.032*(170-T)=8*4.186*T#

#170-T=(8*4.186)/(1*0.032)*T#

#170-T=1046.5T#

#1047.5T=170#

#T=170/1047.5=0.16ºC#

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