An object with a mass of #6 kg#, temperature of #160 ^oC#, and a specific heat of #9 J/(kg*K)# is dropped into a container with #27 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Feb 27, 2017

Answer:

The water does not evaporate, the final temperature is #=0.1ºC#

Explanation:

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=160-T#

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

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

#C_o=9Jkg^-1K^-1#

#m_0 C_o*(160-T) = m_w* 4186 *T#

#6*9*(160-T)=27*4186*T#

#160-T=(27*4186)/(54)*T#

#160-T=2093T#

#2094T=160#

#T=160/2094=0.1ºC#

The water does not evaporate