An object with a mass of #4 kg#, temperature of #261 ^oC#, and a specific heat of #18 (KJ)/(kg*K)# is dropped into a container with #39 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Feb 23, 2017

Answer:

The water does not evapprate and the final temperature is #=79.8#ºC

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=261-T#

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

#C_w=4.186kJkg^-1K^-1#

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

#m_0 C_o*(270-T) = m_w* 4.186 *T#

#4*18*(261-T)=39*4.186*T#

#261-T=(39*4.186)/(72)*T#

#261-T=2.27T#

#3.27T=261#

#T=261/3.27=79.8ºC#

The water does not evaporate