An object with a mass of #2 kg#, temperature of #214 ^oC#, and a specific heat of #33 (KJ)/(kg*K)# is dropped into a container with #26 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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Answer 1 · 2017-12-15T17:17:05

The water does not evaporate and the change in temperature is #=80.8^@C#

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 water is #C_w=4.186kJkg^-1K^-1#

The specific heat of the object is #C_o=33kJkg^-1K^-1#

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

The mass of the water is #m_w=26kg#

#2*33*(214-T)=26*4.186*T#

#214-T=(26*4.186)/(2*33)*T#

#214-T=1.649T#

#2.649T=214#

#T=214/2.649=80.8^@C#

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