An object with a mass of #4 kg#, temperature of #261 ^oC#, and a specific heat of #8 (KJ)/(kg*K)# is dropped into a container with #31 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T07:55:35

The water does not evaporate and the change in temperature is #=51.6ºC#

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=261-T#

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

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

#C_o=8kJkg^-1K^-1#

#4*8*(261-T)=31*4.186*T#

#261-T=(31*4.186)/(4*8)*T#

#261-T=4.0551875T#

#5.0551875T=261#

#T=261/5.0551875~~51.6ºC#

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