An object with a mass of #2 kg#, temperature of #244 ^oC#, and a specific heat of #33 (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

1213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T12:34:55

The water does not evaporate and the temperature changes by #=92.11ºC#

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=244-T#

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

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

#C_o=33kJkg^-1K^-1#

#2*33*(244-T)=26*4.186*T#

#244-T=(26*4.186)/(2*33)*T#

#244-T=1.649T#

#2.649T=244#

#T=244/2.649=92.11ºC#

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