An object with a mass of #1 kg#, temperature of #150 ^oC#, and a specific heat of #24 J/(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
Mar 31, 2018

Answer:

The water does not evaporate and the change in temperature is #=0.06^@C#

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)#

The specific heat of the object is #C_o=0.024kJkg^-1K^-1#

The mass of the object is #m_0=1kg#

The volume of water is #V=15L#

The density of water is #rho=1kgL^-1#

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

#1*0.024*(150-T)=15*4.186*T#

#150-T=(15*4.186)/(1*0.024)*T#

#150-T=2616.3T#

#2617.3T=150#

#T=150/2617.3=0.06^@C#

As the final temperature is #T<100^@C#, the water will not evaporate.