An object with a mass of #2 kg#, temperature of #380 ^oC#, and a specific heat of #25 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?

1 Answer
Mar 8, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.09ºC#

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=380-T#

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

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

#C_o=25Jkg^-1K^-1#

#2*25*(380-T)=48*4186*T#

#380-T=(48*4186)/(50)*T#

#380-T=4018.6T#

#4019.6T=380#

#T=380/4019.6=0.09ºC#

The water does not evaporate