An object with a mass of #3 kg#, temperature of #184 ^oC#, and a specific heat of #37 (KJ)/(kg*K)# is dropped into a container with #15 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Jan 15, 2018

Answer:

The water will evaporate

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=184-T#

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

The specific heat of water is #C_w=4.186kJkg^-1K^-1#

The specific heat of the object is #C_o=37kJkg^-1K^-1#

The mass of the object is #m_0=3kg#

The mass of the water is #m_w=15kg#

#3*37*(184-T)=15*4.186*T#

#184-T=(15*4.186)/(3*37)*T#

#184-T=0.566T#

#1.566T=184#

#T=184/1.566=117.5^@C#

As #T>100^@C#, the water will evaporate