Thermodynamics Focus on Concepts, Question 01 Focus on Concepts, Question 02 Foc

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Thermodynamics Focus on Concepts, Question 01 Focus on Concepts, Question 02 Focus on Concepts, Question 03 Focus on Concepts, Question 04 Focus on Concepts, Question 05 Focus on Concepts, Question 06 Chapter 15, Problem 004 GO Chapter 15, Problem 005 GO Chapter 15, Problem 007 Chapter 15, Problem 008 Review Score Review Results by Study Objective In exercising, a weight lifter loses 0.188 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.53 times 10^5 J. Assuming that the latent heat of vaporization of perspiration is 2.42 times 10^5 1/kg, find the change in the internal energy of the weight after. Determine the minimum number of nutritional calories of food that must be consumed to replace the loss of internal energy. (1 nutritional Calcrie = 4186 J). Number Units Number Units

Explanation / Answer

Q = mL = 0.188 x 2264.76 x 103 = 4.26 x 105J

W = work done = 1.53 x 105 J

U, internal energy change = Q - W = 4.26 x 105 - 1.53 x 105 = 2.73 x 105 J

b)

number of calories = U/4186 = 2.73 x 105 /4186 = 65

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