An object with a mass of #4 kg#, temperature of #281 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #25 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Dec 19, 2017

Answer:

The water does not evaporate and the change in temperature is #=88.4^@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=281-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=12kJkg^-1K^-1#

The mass of the object is #m_0=4kg#

The mass of the water is #m_w=25kg#

#4*12*(281-T)=25*4.186*T#

#281-T=(25*4.186)/(4*12)*T#

#281-T=2.18T#

#3.18T=281#

#T=281/3.18=88.4^@C#

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