An object with a mass of #4 kg#, temperature of #160 ^oC#, and a specific heat of #9 J/(kg*K)# is dropped into a container with #27 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1282 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-20T10:01:45

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=160-T#

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

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

#C_o=0.009kJkg^-1K^-1#

#4*.009*(160-T)=27*4.186*T#

#160-T=(27*4.186)/(4*0.009)*T#

#160-T=3139.5T#

#3140.5T=160#

#T=160/3140.5=0.05ºC#

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