An object with a mass of #4 kg#, temperature of #360 ^oC#, and a specific heat of #6 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 4, 2017

Answer:

The water does not evaporate and the final temperature is #=0.06ºC#

Explanation:

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=360-T#

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

#C_w=4186Jkg^-1K^-1#

#C_o=6Jkg^-1K^-1#

#m_0 C_o*(360-T) = m_w* 4186 *T#

#4*6*(360-T)=32*4186*T#

#360-T=(32*4186)/(24)*T#

#360-T=5581.3T#

#5582.3T=360#

#T=360/5582.3=0.06ºC#

The water does not evaporate