# ch21 - CHAPTER 21 HEAT WORK AND THE FIRST LAW OF...

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Unformatted text preview: CHAPTER 21 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS ActivPhysics can help with these problems: Activities 8.58.13 Section 21-1: The First Law of Thermodynamics Problem 1. In a perfectly insulated container, 1.0 kg of water is stirred vigorously until its temperature rises by 7 . C. How much work was done on the water? Solution Since the container is perfectly insulated thermally, no heat enters or leaves the water in it. Thus, Q = 0 in Equation 21-1. The change in the internal energy of the water is determined from its temperature rise, U = mc T (see comments in Section 19-4 on internal energy), so W =- U =- (1 kg) (4 . 184 kJ / kg K)(7 K) =- 29 . 3 kJ. (The negative sign signifies that work was done on the water.) Problem 2. In a closed but uninsulated container, 500 g of water is shaken violently until its temperature rises by 3 . C. The mechanical work required in the process is 9.0 kJ. (a) How much heat is transferred during the shaking? (b) How much mechanical energy would have been required had the container been perfectly insulated? Solution (a) The change in the internal energy of the water is U = mc T, and the work done by it (i.e., the negative of the work done on it) is given. Therefore, Equation 21-1 gives Q = U + W = (0 . 5 kg) (4 . 184 kJ/kg K)(3 K)+(- 9 . 0 kJ) = 6 . 28 kJ- 9 . 0 kJ =- 2 . 72 kJ. (The negative sign signifies that the water lost heat to its surroundings.) (b) If the water had been in perfect thermal isolation, no heat would have been transferred, Q = 0 and W =- U =- 6 . 28 kJ instead of- 9 . 0 kJ . Problem 3. A 40-W heat source is applied to a gas sample for 25 s, during which time the gas expands and does 750 J of work on its surroundings. By how much does the internal energy of the gas change? Solution Q = 40 W 25 s = 1000 J of heat is added to the gas, which does W = 750 J of work on its surroundings. Thus, the first law of thermodynamics requires that U = Q- W = 1000 J- 750 J = 250 J (an increase in internal energy). Problem 4. What is the rate of heat flow into a system whose internal energy is increasing at the rate of 45 W, given that the system is doing work at the rate of 165 W? Solution From Equation 21-2, dQ/dt = dU/dt + dW/dt = 45 W + 165 W = 210 W . Problem 5. The most efficient large-scale electric power generating systems use high-temperature gas turbines and a so-called combined cycle system that maximizes the conversion of thermal energy into useful work. One such plant produces electrical energy at the rate of 360 MW, while extracting energy from its natural gas fuel at the rate of 670 MW. (a) At what rate does it reject waste heat to the environment? (b) Find its efficiency, defined as the percent of the total energy extracted from the fuel that ends up as work....
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## This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at Berkeley.

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ch21 - CHAPTER 21 HEAT WORK AND THE FIRST LAW OF...

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