ch19 - CHAPTER 19 Heat and the First Law of Thermodynamics...

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CHAPTER 19 Heat and the First Law of Thermodynamics 1 * · Body A has twice the mass and twice the specific heat of body B. If they are supplied with equal amounts of heat, how do the subsequent changes in their temperatures compare? M A = 2 M B ; c A = 2 c B ; C = Mc ; T = Q / C C A = 4 C B ; T A = T B /4 2 · The temperature change of two blocks of masses M A and M B is the same when they absorb equal amounts of heat. It follows that the specific heats are related by ( a ) c A = (M A / M B )c B . ( b ) c A = (M B / M A )c B . ( c ) c A = c B . ( d ) none of the above. ( b ) 3 · The specific heat of aluminum is more than twice that of copper. Identical masses of copper and aluminum, both at 20 ° C, are dropped into a calorimeter containing water at 40 ° C. When thermal equilibrium is reached, ( a ) the aluminum is at a higher temperature than the copper. ( b ) the aluminum has absorbed less energy than the copper. ( c ) the aluminum has absorbed more energy than the copper. ( d ) both ( a ) and ( c ) are correct statements. ( c ) 4 · Sam the shepherd's partner, Bernard, who is a working dog, consumes 2500 kcal of food each day. ( a ) How many joules of energy does Bernard consume each day? ( b ) Sam and Bernard often find themselves sleeping out in the cold night. If the energy consumed by Bernard is dissipated as heat at a steady rate over 24 h, what is his power output in watts as a heater for Sam? (a) E = Q ( b ) P = E / t E = (4.184 × 2.5 × 10 6 ) J = 10.46 MJ P = (10.46 × 10 6 /8.64 × 10 4 ) W = 121 W 5* · A solar home contains 10 5 kg of concrete (specific heat = 1.00 kJ/kg K). How much heat is given off by the concrete when it cools from 25 to 20 ° C?
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Chapter 19 Heat and the First Law of Thermodynamics Q = C T = mc T Q = (10 5 × 10 3 × 5) J = 500 MJ 6 · How many calories must be supplied to 60 g of ice at -10 ° C to melt it and raise the temperature of the water to 40 ° C? Q = C ice T ice + mL f + C water T water Q = 60(10 × 0.49 + 79.7 + 40) cal = 7.48 kcal 7 ·· How much heat must be removed when 100 g of steam at 150 ° C is cooled and frozen into 100 g of ice at 0 ° C? (Take the specific heat of steam to be 2.01 kJ/kg K.) Q = m ( c steam T steam + L v + c w T w + L f ) Q = 100(0.48 × 50 + 540 + 100 + 79.7) cal = 74.4 kcal 8 ·· A 50-g piece of aluminum at 20 ° C is cooled to - 196 ° C by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant and is equal to 0.90 kJ/kg K.) m N L vN = m Al c Al T Al ; solve for m N m N = (50 × 0.9 × 216/199) g = 48.8 g 9* ·· If 500 g of molten lead at 327 ° C is poured into a cavity in a large block of ice at 0 ° C, how much of the ice melts? -
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This note was uploaded on 05/01/2008 for the course PHYS 13/14 taught by Professor Millan/thorstensen during the Spring '08 term at Dartmouth.

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ch19 - CHAPTER 19 Heat and the First Law of Thermodynamics...

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