An object with a mass of #7 kg#, temperature of #235 ^oC#, and a specific heat of #17 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Dec 28, 2017

Answer:

The water will not evaporate and the change in temperature is #=0.21^@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=235-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=0.017kJkg^-1K^-1#

The mass of the object is #m_0=7kg#

The mass of the water is #m_w=32kg#

#7*0.017*(235-T)=32*4.186*T#

#235-T=(32*4.186)/(7*0.017)*T#

#235-T=1125.6T#

#1126.6T=235#

#T=235/1126.6=0.21^@C#

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