An object with a mass of #6 kg#, temperature of #50 ^oC#, and a specific heat of #19 J/(kg*K)# is dropped into a container with #24 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-09T14:40:34

The water does not evaporate and the change in temperature is #=0.06^@C#

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=50-T#

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

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

#C_o=0.019kJkg^-1K^-1#

#6*0.019*(50-T)=24*4.186*T#

#50-T=(24*4.186)/(6*0.019)*T#

#50-T=881.3T#

#882.3T=50#

#T=50/882.3=0.06^@C#

As #T<100^@C#, the water does not evaporate