An object with a mass of #2 kg#, temperature of #50 ^oC#, and a specific heat of #19 J/(kg*K)# is dropped into a container with #24 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Feb 28, 2017

Answer:

The water does not evaporate. The change in temperature is #=0.16º#C

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=50-T#

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

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

#C_o=19Jkg^-1K^-1#

#m_0 C_o*(50-T) = m_w* 4186 *T#

#2*19*(50-T)=24*4186*T#

#50-T=(24*4186)/(38)*T#

#50-T=306.9T#

#307.9T=50#

#T=50/307.9=0.16ºC#

The water does not evaporate