An object with a mass of #1 kg#, temperature of #125 ^oC#, and a specific heat of #32 (KJ)/(kg*K)# is dropped into a container with #15 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 · 2017-06-04T06:36:10

The water does not evaporate and the change in temperature is #=42.2ºC#

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=125-T#

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

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

#C_o=32kJkg^-1K^-1#

#1*32*(125-T)=15*4.186*T#

#125-T=(15*4.186)/(1*32)*T#

#125-T=1.96T#

#2.96T=125#

#T=125/2.96=42.2ºC#

As #T<100ºC#, the water does not evaporate