Origen

Chapter 17. Quantity of Heat

NOTE: Refer to Tables 18-1 and 18-2 for accepted values for specific heat, heat of vaporization, and heat of fusion for the substances in the problems below

Quantity of heat and Specific Heat Capacity
17-1. What quantity of heat is required to change the temperature of 200 g of lead from 20 to 1000C? [Answer given in calories--also worked below for joules.][pic]; Q = 2080 J
[pic]; Q = 496 cal

17-2. A certain process requires 500 J of heat. Express this energy in calories and in Btu.
[pic] Q = 119 cal
[pic] Q = 0.474 Btu

17-3. An oven applies 400 kJ of heat to a 4 kg of a substance causing its temperature to increase by 80 C0. What is the specific heat capacity?
[pic]; c = 1250 J/kg C0
17-4. What quantity ofheat will be released when 40 lb of copper cools form 78 to 320F?
Q = (40 lb)(0.093 Btu/lb F0)(780F – 320F); Q = 171 Btu

17-5. A lawn mower engine does work at the rate of 3 kW. What equivalent amount of heat will be given off in one hour?
[pic]; Q = 10,800,000 J/h; Q = 10.8 MJ/h
17-6. An air conditioner is rated at 15,000 Btu/h. Express this power in kilowatts and in caloriesper second?
[pic]; P = 1050 cal/s
17-7. Hot coffee is poured into a 0.5-kg ceramic cup with a specific heat of 880 J/kg C0. How much heat is absorbed by the cup if it’s temperature increases from 20 to 800C?
Q = mcΔt = (0.5 kg)(880 J/kg C0)(800C – 200C); Q = 26.4 kJ

17-8. A 2 kW electric motor is 80 percent efficient. How much heat is lost in one hour?
Loss = 0.20(2 kW) = 400 W= 400 J/s;
Loss = 400 J/s(3600 s/h); Loss = 1.44 MJ

17-9. An 8-kg copper sleeve must be heated from 250C to 1400C in order that it will expand to fit over a shaft. How much heat is required?
Q = mcΔt = (8 kg)(390 J/kg C0)(1400C – 250C); Q = 359 kJ.

17-10. How many grams of iron at 200C must be heated to 1000C in order to be able to release 1800 cal of heat as it returns toits original temperature?
[pic]; m = 199 g

17-11. A 4-kg chunk of metal (c = 320 J/kg C0) is initially at 3000C. What will its final temperature be if it losses 50 kJ of heat energy?
[pic]; Δt = -39.1 C0
t = 3000C – 39.10C; t = 2610C
17-12. In a heat-treatment, a hot copper part is quenched with water, cooling it from 4000C to 300C. What was the mass of the part if itloses 80 kcal of heat?
[pic]; m = 2325 g; m = 2.32 kg

Conservation of Energy: Calorimetry
*17-13. A 400-g copper pipe initially at 2000C is dropped into a container filled with 3 kg of water at 200C. Ignoring other heat exchanges, what is the equilibrium temperature of the mixture?
Heat lost by copper pipe = Heat gained by water; mccc Δt = mwcw Δt;
(400 g)(0.093 cal/g C)(2000C– te) = (3000 kg)(1 ca/g C)(te – 200C)
Solving for te we obtain: te = 22.20C

17-14. How much aluminum (c = 0.22 cal/g C0) at 200C must be added to 400 g of hot water at 800C in order that the equilibrium temperature be 300C?
mw cw (800C – 300C) = mAL cAl (300C – 200C)
(400 g)(1 cal/g C0)(50 C0) = mAl (0.22 cal/g C0)(10 C0)
[pic]; mAl = 9.09 kg

17-15. A 450-g chunk of metal isheated to 1000C and then dropped into a 50-g aluminum calorimeter cup containing 100-g of water. The initial temperature of cup and water is 100C and the equilibrium temperature is 21.10C. Find the specific heat of the metal?
Heat lost by metal = heat gained by cup + heat gained by water
mxcx(1000C – 21.10C) = mAlcAl(21.10C – 100C) + mwcw(21.10C – 100C)
17-15. (Cont.)mxcx(78.9 C0) = mAlcAl(11.1 C0) + mwcw(11.1 C0)
(450 g) cx (78.9 C0) = (50 g)(0.22 cal/g C0)(11.1 C0) + (100 g) (1 cal/g C0)(11.1 C0)
(35,505 g C0) cx = 122.1 cal + 1110 cal; cx = 0.0347 cal/g C0

17-16. What mass of water initially at 200C must be mixed with 2 kg of iron to bring the iron from 2500C to an equilibrium temperature of 250C?
mw cw (250C – 200C) = miron ciron (2500C –...