An object with a mass of #2 kg#, temperature of #320 ^oC#, and a specific heat of #25 J/(kg*K)# is dropped into a container with #16 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1348 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-19T06:28:09

The water does not evaporate and the change in temperature is #=0.25ºC#

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=320-T#

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

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

#C_o=0.025kJkg^-1K^-1#

#2*.025*(320-T)=16*4.186*T#

#320-T=(16*4.186)/(2*0.025)*T#

#320-T=1339.5T#

#1340.5T=320#

#T=320/1340.5=0.25ºC#

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