An object with a mass of #12 kg#, temperature of #210 ^oC#, and a specific heat of #9 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 3, 2017

Answer:

The water does not evaporate and the final temperature is #=0.17ºC#

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=210-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*(210-T) = m_w* 4186 *T#

#12*9*(210-T)=32*4186*T#

#210-T=(32*4186)/(108)*T#

#210-T=1240.3T#

#1241.3T=210#

#T=210/1241.3=0.17ºC#

The water does not evaporate