An object with a mass of #12 kg#, temperature of #210 ^oC#, and a specific heat of #7 J/(kg*K)# is dropped into a container with #48 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1289 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-07T13:48:51

The water does not evaporate and the change in temperature is negligible.

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=210-T#

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

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

#C_o=0.007kJkg^-1K^-1#

#m_0=12kg#

#12*0.007*(210-T)=48*4.186*T#

#210-T=(48*4.186)/(12*0.007)*T#

#210-T=2392T#

#2393T=210#

#T=210/2393=0.00004^@C#

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