General Physics II Exam 1-Page 4 of 9 02/28/2018 3. (10 points) A block of ice w

Question

General Physics II Exam 1-Page 4 of 9 02/28/2018 3. (10 points) A block of ice whose mass is 50.0g (at 0'C) is mixed with 200.0g of water that is initially at 75 C. The heat capacity for water is 4190x and the heat of fusiou for ice is LF = 33 (a) (3 points) What is the final temperature of the water when the system reaches thermal equilibrium? (b) (2 points) What is the change in the entropy associated with the ice melting? (e) (2 points) What is the change in the entropy of the melted ice water when it changes temperature from zero to the equilibrium temperature? (d) (2 points) What is the change in the entropy of the hot water when it changes temperature from 75 C to the equilibrium temperature? (e) (1 point) What is the change in the entropy of the systen?

Explanation / Answer

a)Let the final temperature of water be T.so,

Heat energy lost by water at 75 C = heat energy gained by water at at 0C + heat gained on melting.

so, conserving energy,

200*4.19*(75-T) = 50*4.19*T + 50*334

or T = 44.05 C

b)Change in Entropy for melting of ice = heat enrgy gained by ice/temperature

= 50*334/(273.15) (temperature has to be in Kelvin for calculating entropy)

=61.1386 J/K

c)Entropy change on warming of water from 0C to 44.05C = C*ln(Tf/Ti)

=50*4.19*ln((273.15+40.05)/(273.15))

= 28.664 J/K

d)Entropy of colling of 200g of water from 75 to 44.05C = C*ln(Tf/Ti)

=200*4.19*ln((273.15+40.05)/(273+75))

=-88.2921 J/K

e)total change in entropy of the system = -88.2921+28.664+61.1368

=1.5086 J/K

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