An object with a mass of #3 kg#, temperature of #210 ^oC#, and a specific heat of #24 J/(kg*K)# is dropped into a container with #36 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 · 2017-12-23T06:44:55

The water will not evaporate and the change in temperature is #=0.1^@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=210-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=0.024kJkg^-1K^-1#

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

The mass of the water is #m_w=36kg#

#3*0.024*(210-T)=36*4.186*T#

#210-T=(36*4.186)/(3*0.024)*T#

#210-T=2093T#

#2094T=210#

#T=210/2094=0.1^@C#

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