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PART 3 THERMODYNAMICS CHAPTER 19 TEMPERATURE AND HEAT --- - - - - Problem Problem 3. Normal room temperature is 68OF. What is this in Celsius? Solution Equation 19-3, solved for the Celsius temperature, gives Tc = i(TF - 32) = 5(68 - 32)/9 = 20°C. Sections 19-4 and 19-5: Temperature and Heat, Heat Capacity and Specific Heat Problem 14. The average human diet contains 2000 kcal per day. If all this food energy is released rather than being stored as fat, what is the approximate average power output of the human body? Solution pa, = (2x10~ cal/d)(4.184 J/cal)(l d/86,400 S) = 96.9 W. Problem 19. A circular lake 1.0 krn in diameter averages 10 m deep. Solar energy is incident on the lake at an average rate of 200 W/m2. If the lake absorbs all this energy and does not exchange heat with its surroundings, how long will it take to warm from 10°C to 20°C? 27. A stove burner supplies heat at the rate of 1.0 kW, a microwave oven at 625 W. You can heat water in the microwave in a paper cup of negligible heat capacity, but the stove requires a pan whose heat capacity is 1.4 kJ/K. (a) How much water do you need before it becomes quicker to heat on the stovetop? (b) What will be the rate at which the temperzture of this much water rises? Solution The temperature rise per second is equal to the heat supplied per second (i.e., the power supplied if there are no losses) divided by the total heat capacity of the water and its container: AT/At = (AQ/At)/Cto, = P/Ctot (Equation 19-4 divided by At). The total heat capacity is Ctot = CW + Cent, provided the water and container both have the same instantaneous temperature. (This assumes that heat is supplied sufficiently slowly that the water and container share it and stay in instantaneous thermal equilibrium.) For the paper cup used in the microwave oven, C,,, x 0, whereas for the pan used on the stove burner, Cent = 1.4 kJ/K. Thus, the rate of temperature rise is - 625 W/Cw for the microwave and 1.0 kW/(Cw + 1.4 kJ/K) for the stove burner. (a) When = mwcw is small, the microwave is faster, whereas when Cw is large, the stove burner is faster. (To see this, plot both rates as a function of Cw.) The rates of temperature rise are equal for Cw = mw x (4.184 kJ1kg.K) = (1.4 kJ/K)(0.625)/(1 - 0.625) = 2.33 kJ/K. Therefore, mw = (2.33 kJ/K)+ Solution (4.184 kJ/kg-K) = 0.558 kg. (b) For mw in part (a), Since the energy absorbed by ttw lake equals the solar the rate of temperature rise is ATlAt = (0.625 kW)s power times the time, At = AQ/P = ,CAT+ (2.33 kJ/K) = 1.0 kWl(2.33
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