An object with a mass of #6 kg#, temperature of #315 ^oC#, and a specific heat of #12 (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

1177 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-26T09:02:51

The final temperature is #=100ºC#
The water does not evaporate.

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=315-T#

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

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

#C_o=12kJkg^-1K^-1#

#m_0 C_o*(315-T) = m_w* 4.186 *T#

#6*12*(315-T)=37*4.186*T#

#315-T=(37*4.186)/(72)*T#

#315-T=2.15T#

#3.15T=315#

#T=315/3.15=100ºC#

The water does not evaporate