An object with a mass of #3 kg#, temperature of #270 ^oC#, and a specific heat of #18 J/(kg*K)# is dropped into a container with #36 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Feb 20, 2017

Answer:

The water does not evaporate. The increase in temperature of the water #=0.1ºC#

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by 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=270-T#

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

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

#C_o=18Jkg^-1K^-1#

#m_0 C_o*(270-T) = m_w* 4186 *T#

#3*18*(270-T)=36*4186*T#

#270-T=(36*4186)/(3*18)*T#

#270-T=2790.7T#

#2791.7T=270#

#T=270/2791.7=0.1ºC#