An object with a mass of #2 kg#, temperature of #214 ^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?

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Mar 28, 2018

Answer:

The water will not evaporate and the change in temperature is #=39.9^@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=214-T#

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

The specific heat of the object is #C_o=12kJkg^-1K^-1#

The mass of the object is #m_0=2kg#

The volume of water is #V=25L#

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

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

#2*12*(214-T)=25*4.186*T#

#214-T=(25*4.186)/(2*12)*T#

#214-T=4.36T#

#5.36=214#

#T=214/5.36=39.9^@C#

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

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