An object with a mass of #2 kg#, temperature of #150 ^oC#, and a specific heat of #24 J/(kg*K)# is dropped into a container with #18 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 4, 2017

Answer:

The water does not evaporate and the change in temperature is #0.01ºC#

Explanation:

The heat is transferred from the hot object to 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=150-T#

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

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

#C_o=24Jkg^-1K^-1#

#m_0 C_o*(150-T) = m_w* 4186 *T#

#2*24*(150-T)=18*4186*T#

#150-T=(18*4186)/(48)*T#

#150-T=1569.75T#

#1570.75T=150#

#T=150/1570.75=0.01ºC#

The water does not evaporate