An object with a mass of #2 kg#, temperature of #150 ^oC#, and a specific heat of #24 J/(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 · 2017-08-19T07:53:57

The water doesn't evaporate

#V = 15L=0.015m^(-3)#
#rho = 1000kg*m^(-3)#

#rho=m/V#

#m = 0.015*1000#
#rArr m = 15kg#

Let final temperature of object and water be #T_f#

According to the principle of calorimetry,

#Q_o = Q_w#
#2*24*(150-T_f) = 15*4.2*T_f#
#rArr T_f ~~64.8^o C#

#:. # The water wouldn't evaporate and #DeltaT ~~ 65^oC#