An object with a mass of #18 kg#, temperature of #240 ^oC#, and a specific heat of #6 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
Feb 24, 2017

Answer:

The water does not evaporate and the rise in temperature ot the water is #=0.13ºC#

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=240-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*(240-T) = m_w* 4186 *T#

#18*6*(240-T)=48*4186*T#

#240-T=(48*4186)/(108)*T#

#240-T=1860.4T#

#1861.4T=240#

#T=240/1861.4=0.13ºC#

The water does not evaporate