An object with a mass of #2 kg#, temperature of #331 ^oC#, and a specific heat of #18 (KJ)/(kg*K)# is dropped into a container with #37 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1062 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-29T06:33:33

The water does not evaporate and the change in temperature is #=0.08ºC#

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=331-T#

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

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

#C_o=0.018kJkg^-1K^-1#

#2*0.018*(331-T)=37*4.186*T#

#331-T=(37*4.186)/(2*0.018)*T#

#331-T=4302.3T#

#4303.3T=331#

#T=331/4303.3=0.08ºC#

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