An object with a mass of #2 kg#, temperature of #214 ^oC#, and a specific heat of #13 (KJ)/(kg*K)# is dropped into a container with #26 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1151 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-06T06:30:44

The water does not evaporate and the change in temperature is #=0.05ºC#

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=214-T#

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

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

#C_o=0.013kJkg^-1K^-1#

#2*0.013*(214-T)=26*4.186*T#

#214-T=(26*4.186)/(2*0.013)*T#

#214-T=4186T#

#4187T=214#

#T=214/4187=0.05ºC#

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