(a)
The entropy rise of the entire system.
(a)
Answer to Problem 56P
The entropy rise of the entire system is 13.4 J/K.
Explanation of Solution
Given info: The mass of the athlete and the water is 70 kg and 454 g respectively. The initial temperature of athlete and water is 98.6 °F and 35 °F respectively.
Write the expression to calculate the change in entropy of the system.
ΔS=ΔSice water+ΔSbody (1)
Here,
ΔSice water is the change in the entropy of cold water.
ΔSbody is the change in the entropy of body.
ΔS is change in entropy of the system.
Write the expression to calculate the change in entropy of water.
ΔSice water=ms∫dTT (2)
Here,
m is the mass of water.
s is the specific heat of water.
T is the absolute temperature.
dT is the change in temperature of water.
Write the expression to convert the temperature from Fahrenheit to Kelvin.
K=59(°F−32)+273.15 (3)
Substitute 98.6 °F for °F in equation (3).
K=59(98.6 °F−32)+273.15=310.15 K
Thus, the temperature of body in Kelvin is 310.15 K.
Substitute 35 °F for °F in equation (3).
K=59(35 °F-32)+273.15=274.82 K
Thus, the temperature of water in Kelvin is 274.82 K.
Substitute 454 g for m, 4.18 J/g⋅K for s in equation (1) to find ΔSice water.
ΔSice water=454 g(4.18 J/g⋅K)∫dTT=1897.72 J/K∫dTT
Integrate the above expression from the limit of 274.67 K to 310 K.
ΔSice water=1897.72 J/K310 K∫274.67 KdTT
Write the expression to calculate the change in entropy of water.
ΔSbody=−ms(T2−T1)T2
Here,
T2 is the temperature of body.
T1 is the temperature of water.
Substitute −msdTT for ΔSbody and 1897.72 J/K310 K∫274.67 KdTT for ΔSice water in equation (1).
ΔS=1897.72 J/K310 K∫274.67 KdTT+−ms(T2−T1)T2 (4)
Substitute 454 g for m, 4.18 J/g⋅K for s, 310 K for T2 and 274.67 K for T1 in equation (4) to find ΔS.
ΔS=1897.72 J/K310 K∫274.67 KdTT−454 g×4.18 J/g⋅K(310 K−274.67 K)310 K=1897.72 J/K×ln(310 K274.67 K)−216.27 J/K=13.4 J/K
Thus, the entropy rise of the entire system is 13.4 J/K.
Conclusion:
Therefore, the entropy rise of the entire system is 13.4 J/K.
(b)
The athlete’s temperature after she drinks the cold water.
(b)
Answer to Problem 56P
The final temperature of the body is 310 K.
Explanation of Solution
Given info: The mass of the athlete and the water is 70 kg and 454 g respectively. The initial temperature of athlete and water is 98.6 °F and 35 °F respectively.
Write the expression of heat balance equation.
Heat gained by water=Heat lost by bodyms(Tf−274.82 K)=Ms(310.15 K−Tf)m(Tf−274.82 K)=M(310.15 K−Tf) (5)
Here,
Tf is the final temperature of the body.
Substitute 454 g for m and 70 kg for M in equation (5).
454 g(Tf−274.82 K)=70 kg×1000 g1 kg(310.15 K-Tf)Tf=309.92 K≈310 K
Conclusion:
Therefore, the final temperature of the body is 310 K.
(c)
The entropy rise of the entire system.
(c)
Answer to Problem 56P
The entropy rise of the entire system is 13.3 J/K.
Explanation of Solution
Given info: The mass of the athlete and the water is 70 kg and 454 g respectively. The initial temperature of athlete and water is 98.6 °F and 35 °F respectively.
Write the expression to calculate the change in entropy of the system.
ΔS=ΔSice water+ΔSbody (1)
Write the expression to calculate the change in entropy of water.
ΔSice water=ms∫dTT (2)
Integrate the above expression from the limit of 274.67 K to 309.78 K.
ΔSice water=ms309.92 K∫274.82 KdTT
Substitute 454 g for m, (1 cal/g⋅K) for s in above equation.
ΔSice water=454 g(1 cal/g⋅K)309.92 K∫274.82 KdTT=54.6 cal/K≈55 cal/K
Write the expression to calculate the change in entropy of body.
ΔSbody=Ms∫dTT (7)
Here,
M is the mass of athlete.
Integrate the above expression from the limit of 309.92 K to 310.15 K.
ΔSbody=Ms309.92 K∫310.15 KdTT
Substitute 70 kg for M and (1 cal/g⋅K) for s in above equation.
ΔSbody=70 kg×1000 g1 kg(1 cal/g⋅K)309.92 K∫310.15 KdTT=−51.8 cal/K
Substitute 55 cal/K for ΔSice water and −51.8 cal/K for ΔSbody in equation (1) to find ΔS.
` ΔS=55 cal/K−51.8 cal/K=3.2 cal/K=3.2 cal/K×4.18 J1 cal≈13.3 J/K (8)
Thus, the entropy rise of the entire system is 13.3 J/K.
Conclusion:
Therefore, the entropy rise of the entire system is 13.3 J/K.
(d)
The result by comparing the part (a) and (c).
(d)
Answer to Problem 56P
The change in entropy in part (c) is less than that of part (a) by less than 1%.
Explanation of Solution
Given info: The mass of the athlete and the water is 70 kg and 454 g respectively. The initial temperature of athlete and water is 98.6 °F and 35 °F respectively.
The percentage change in entropy is,
%S=13.4 J/K−13.3 J/K13.4 J/K×100=0.74%
Thus the change in entropy in part (c) is less than that of part (a) by less than 1%.
Conclusion:
Therefore, the change in entropy in part (c) is less than that of part (a) by less than 1%.
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Chapter 18 Solutions
Principles of Physics: A Calculus-Based Text
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