An object with a mass of #3 kg#, temperature of #331 ^oC#, and a specific heat of #18 (KJ)/(kg*K)# is dropped into a container with #42 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Feb 5, 2018

Answer:

The water does not evaporate and the change in temperature is #=77.8^@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=331-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=18kJkg^-1K^-1#

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

The volume of water is #V=42L#

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

The mass of the water is #m_w=rhoV=42kg#

#3*18*(331-T)=42*4.186*T#

#331-T=(42*4.186)/(3*18)*T#

#331-T=3.26T#

#4.26T=331#

#T=331/4.26=77.8^@C#

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