# Chapter 17. Quantity of Heat. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

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1 Chapter 17. Quantity of Heat A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 Photo Vol. 05 Photodisk/Getty FOUNDRY: It requires about 289 Joules of heat to melt one gram of steel. In this chapter, we will define the quantity of heat to raise the temperature and to change the phase of a substance.

3 Objectives: After finishing this unit, you should be able to: Define the quantity of heat in terms of the calorie, the kilocalorie, the joule, and the Btu. Write and apply formulas for specific heat capacity and solve for gains and losses of heat. Write and apply formulas for calculating the latent heats of fusion and vaporization of various materials.

4 Heat Defined as Energy Heat is is not something an object has, but rather energy that it it absorbs or or gives up. The heat lost by the hot coals is is equal to to that gained by the water. Cool water Thermal Equilibrium Hot coals

5 Units of Heat One calorie (1 cal) is the quantity of heat required to raise the temperature of 1 g of water by 1 C 0. Example 10 calories of heat will raise the temperature of 10 g of water by 10 C 0.

6 Units of Heat (Cont.) One kilocalorie (1 kcal) is the quantity of heat required to raise the temperature of 1 kg of water by 1 C 0. Example 10 kilocalories of heat will raise the temperature of 10 kg of water by 10 C 0.

7 Units of Heat (Cont.) One British Thermal Unit (1 Btu) is the quantity of heat required to raise the temperature of 1 lb of water by 1 F 0. Example 10 Btu of heat will raise the temperature of 10 lb of water by 10 F 0.

8 The Btu is an Outdated Unit The British Thermal Unit (1 Btu) is discouraged, but unfortunately remains in wide-spread use today. If it is to be used, we must recognize that the pound unit is actually a unit of mass, not weight. 1 lb 1 lb (1/32) slug When working with the Btu, we must recall that the pound-mass is not a variable quantity that depends on gravity -- one reason that the use of the Btu is discouraged!

9 The SI Unit of Heat Since heat is is energy, the joule is is the preferred unit. Then, mechanical energy and heat are measured in in the same fundamental unit. Comparisons of Heat Units: 1 cal = J 1 Btu = 778 ft ft lb lb 1 kcal = 4186 J 1 Btu = 252 cal 1 Btu = 1055 J

10 Temperature and Quantity of Heat The effect of heat on temperature depends on the quantity of matter heated. The same quantity of heat is applied to each mass of water in the figure. The larger mass experiences a smaller increase in temperature C 600 g 20 0 C 200 g 22 0 C 30 0 C

11 Heat Capacity The heat capacity of a substance is the heat required to raise the temperature a unit degree. Lead Glass Al Copper Iron C C C C C 37 s 52 s 60 s 83 s 90 s Heat capacities based on time to heat from zero to C. Which has the greatest heat capacity?

12 Heat Capacity (Continued) All at at C placed on Paraffin Slab Lead Glass Al Copper Iron Iron and copper balls melt all the way through; others have lesser heat capacities.

13 Specific Heat Capacity The specific heat capacity of of a material is is the quantity of of heat needed to to raise the temperature of of a unit mass through a unit degree. Q c ; Q mc t m t Water: c = 1.0 cal/g C 0 or or 1 Btu/lb F 0 or or 4186 J/kg K Copper: c = cal/g C 0 or or 390 J/kg K

14 Comparison of Heat Units: How much heat is needed to raise 1-kg 1 of water from 0 0 to C? The mass of one kg of water is: 1 kg = 1000 g = lb m Q mc t 1 lb m = 454 g For water: c = 1.0 cal/g C 0 or 1 Btu/lb F 0 or 4186 J/kg K The heat required to do this job is: 10,000 cal 10 kcal 39.7 Btu 41, 860 J 1 kg

15 Problem Solving Procedure 1. Read problem carefully and draw a rough sketch. 2. Make a list of all given quantities 3. Determine what is to be found. 4. Recall applicable law or formula and constants. Q c ; Q mc t m t Water: c = cal/g C 00 or or 1 Btu/lb F 00 or or 4186 J/kg K 5. Determine what was to be found.

16 Example 1: A 500-g copper coffee mug is filled with 200-g of coffee. How much heat was required to heat cup and coffee from 20 to 96 0 C? 1. Draw sketch of problem. 2. List given information. Mug mass m m = kg Coffee mass m c = kg Initial temperature of coffee and mug: t 0 = 20 0 C Final temperature of coffee and mug: t f = 96 0 C 3. List what is to be found: Total heat to raise temperature of coffee (water) and mug to 96 0 C.

17 Example 1(Cont.): How much heat needed to heat cup and coffee from 20 to 96 0 C? m m = 0.2 kg; ; m w = 0.5 kg. 4. Recall applicable formula or law: Heat Gain or Loss: Q = mc t 5. Decide that TOTAL heat is that required to raise temperature of mug and water (coffee). Write equation. Q T = m m c m t + m w c w t 6. Look up specific heats in tables: Copper: c m = 390 J/kg C 0 Coffee (water): c w = 4186 J/kg C 0

18 Example 1(Cont.): How much heat needed to heat cup and coffee from 20 to 96 0 C? m c = 0.2 kg; ; m w = 0.5 kg. 7. Substitute info and solve problem: Copper: c m = 390 J/kg C 0 Coffee (water): c w = 4186 J/kg C 0 Q T =m m c m t+ m w c w t Water: (0.20 kg)(4186 J/kgC 0 )(76 C 0 ) Cup: (0.50 kg)(390 J/kgC 0 )(76 C 0 ) t = 96 0 C C = 76 C 0 Q T = 63,600 J + 14,800 J Q T = 78.4 kj kj

19 A Word About Units The substituted units must be consistent with those of the chosen value of specific heat capacity. For example: Water c w = 4186 J/kg C 0 or 1 cal/g C 0 Q = m w c w t If If The you units use for cal/g Q, J/kg Cm, 0 Cfor and 0 for c, c, t c, c, then must Q be must consistent be in in calories, joules, with and m and those must m must based be in in be kilograms. on in in the value of the constant c.

20 Conservation of Energy Whenever there is a transfer of heat within a system, the heat lost by the warmer bodies must equal the heat gained by the cooler bodies: (Heat Losses) = (Heat Gained) Hot iron Cool water Thermal Equilibrium

21 Example 2: A handful of copper shot is heated to 90 0 C and then dropped into 80 g of water in an insulated cup at 10 0 C.. If the equilibrium temperature is 18 0 C, what was the mass of the copper? c w = 4186 J/kg C 0 ; c s = 390 J/kg C shot Insulator t e = 18 0 C 10 0 water m w = 80 g; t w = 10 0 C; t s = 90 0 C Heat lost by shot = heat gained by water m s c s (90 0 C C) = m w c w (18 0 C C) Note: Temperature differences are [High - Low] to insure absolute values (+) lost and gained.

22 m s =? 90 0 shot Insulator Example 2: (Cont.) 10 0 water 18 0 C 80 g of Water Heat lost by shot = heat gained by water m s c s (90 0 C C) = m w c w (18 0 C C) m s (390 J/kgC 0 )(72 C 0 ) = (0.080 kg)(4186 J/kgC 0 )(8 C 0 ) 2679 J ms kg m s 28,080 J/kg s = 95.4 g

23 Change of Phase When a change of of phase occurs, there is is only a change in in potential energy of of the molecules. The temperature is is constant during the change. Solid fusion Liquid Vaporization Gas Q = ml f Q = ml v Terms: Fusion, vaporization, condensation, latent heats, evaporation, freezing point, melting point.

24 Change of Phase The latent heat of fusion (L f ) of a substance is the heat per unit mass required to change the substance from the solid to the liquid phase of its melting temperature. For Water: L f f = cal/g = 333,000 J/kg L f Q m The latent heat of vaporization (L v ) of a substance is the heat per unit mass required to change the substance from a liquid to a vapor at its boiling temperature. L v Q m For Water: L v v = 540 cal/g = 2,256,000 J/kg

25 Melting a Cube of Copper The heat Q required to melt a substance at its melting temperature can be found if the mass and latent heat of fusion are known. 2 kg What Q to melt copper? Q = ml vv L f = 134 kj/kg Example: To completely melt 2 kg of copper at C, we need: Q = ml f = (2 kg)(134,000 J/kg) Q = 268 kj kj

26 Example 3: How much heat is needed to convert 10 g of ice at C to steam at C? First, let s review the process graphically as shown: temperature t ice steam 540 cal/g C 0 0 C C 1 cal/gc 0 80 cal/g water ice c ice and = 0.5 cal/gc only 0 ice water steam and water steam only Q

27 Example 3 (Cont.): Step one is Q 1 to convert 10 g of ice at C to ice at 0 0 C (no water yet) C 0 0 C t C Q 1 to raise ice to 0 0 C: Q 1 = mc t Q 1 = (10 g)(0.5 cal/gc 0 )[0 - (-20 0 C)] Q 1 = (10 g)(0.5 cal/gc 0 )(20 C 0 ) 0 0 C C ice c ice = 0.5 cal/gc 0 Q 1 = 100 cal Q

28 Example 3 (Cont.): Step two is Q 2 to convert 10 g of ice at 0 0 C to water at 0 0 C. Melting t C Q 2 to melt 10 g of ice at 0 0 C: Q 2 = ml f Q 2 = (10 g)(80 cal/g) = 800 cal Q 2 = 800 cal 0 0 C C 80 cal/g ice and water Add this to Q 1 = 100 cal: 900 cal used to this point. Q

29 Example 3 (Cont.): Step three is Q 3 to change 10 g of water at 0 0 C to water at C. 0 0 C to C t C Q 3 to raise water at 0 0 C to C. Q 3 = mc t ; c w = 1 cal/gc 0 Q 3 = (10 g)(1 cal/gc 0 )(100 0 C C) 1 cal/gc 0 Q 3 = 1000 cal 0 0 C C water only Total = Q 1 + Q 2 + Q 3 = = 1900 cal Q

30 Example 3 (Cont.): Step four is Q 4 to convert 10 g of water to steam at C?? ( (Q 4 = ml v ) vaporization Q 4 to convert all water at C to steam at C. (Q = ml v ) C Q 4 = (10 g)(540 cal/g) = 5400 cal 0 0 C C 800 cal 100 cal ice ice and water 1000 cal water only 5400 cal Total Heat: steam and water 7300 cal Q

31 Example 4: How many grams of ice at 0 0 C must be mixed with four grams of steam in order to produce water at 60 0 C? m i =? Ice must melt and then rise to 60 0 C. Steam must condense and drop to 60 0 ice C. steam Total Heat Gained = Total Heat Lost 4 g m i L f + m i c w t= m s L v + m s c w t t e = 60 0 C Note: All losses and gains are absolute values (positive). Total Gained: m i i -0 0 i (80 cal/g) + m i (1 cal/gc 0 )(60 C 0 ) C) Total Lost: Lost: (4 g)(540 (4 g)(540 cal/g) cal/g) + (4 g)(1 + (4 cal/gc g)(1 cal/gc 0 )(100 0 )(40 C 0-60 C 00 ) C)

32 Example 4 (Continued) Total Gained: m i (80 cal/g) + m i (1 cal/gc 0 )(60 C 0 ) Total Lost: (4 g)(540 cal/g) + (4 g)(1 cal/gc 0 )(40 C 0 ) Total Heat Gained = Total Heat Lost 80m i + 60m i = 2160 g +160 g m i =? 4 g mi 2320 g 140 m i i = 16.6 g t e = 60 0 C

33 Example 5: Fifty grams of ice are mixed with 200 g of water initially at 70 0 C.. Find the equilibrium temperature of the mixture. Ice melts and rises to t e Water drops from 70 to t e. ice water 0 0 C 70 0 C 50 g 200 g t e =? Heat Gained: m i L f + m i c w t ; t = t e -0 0 C Gain = (50 g)(80 cal/g) + (50 g)(1 cal/gc 0 )(t e -0 0 C) Gain = 4000 cal + (50 cal/g)t e

34 Example 5 (Cont.): Gain = 4000 cal + (50 cal/g)t e Heat Lost = m w c w t t = 70 0 C -t e [high - low] 0 0 C 70 0 C 50 g 200 g t e =? Lost = (200 g)(1 cal/gc 0 )(70 0 C- t e ) Lost = 14,000 cal - (200 cal/c 0 ) t e Heat Gained Must Equal the Heat Lost: 4000 cal + (50 cal/g)t e = 14,000 cal - (200 cal/c 0 ) t e

35 Example 5 (Cont.): Heat Gained Must Equal the Heat Lost: 4000 cal + (50 cal/g)t e = 14,000 cal - (200 cal/c 0 ) t e Simplifying, we have: (250 cal/c 0 ) t e = 10,000 cal t e 10,000 cal 250 cal/c C 0 0 C 70 0 C 50 g 200 g tt e e = 40 0 C t e =?

36 Summary of Heat Units One calorie (1 cal) is the quantity of heat required to raise the temperature of 1 g of water by 1 C 0. One kilocalorie (1 kcal) is the quantity of heat required to raise the temperature of 1 kg of water by 1 C 0. One British thermal unit (Btu) is the quantity of heat required to raise the temperature of 1 lb of water by 1 F 0.

37 Summary: Change of Phase The latent heat of fusion (L f ) of a substance is the heat per unit mass required to change the substance from the solid to the liquid phase of its melting temperature. For Water: L f f = cal/g = 333,000 J/kg L f Q m The latent heat of vaporization (L v ) of a substance is the heat per unit mass required to change the substance from a liquid to a vapor at its boiling temperature. L v Q m For Water: L v v = 540 cal/g = 2,256,000 J/kg

38 Summary: Specific Heat Capacity The specific heat capacity of of a material is is the quantity of of heat to to raise the temperature of of a unit mass through a unit degree. Q c ; Q mc t m t

39 Summary: Conservation of Energy Whenever there is a transfer of heat within a system, the heat lost by the warmer bodies must equal the heat gained by the cooler bodies: (Heat Losses) = (Heat Gained)

40 Summary of Formulas: Q c ; Q mc t m t (Heat Losses) = (Heat Gained) Q Lf ; Q ml m Q Lv ; Q ml m f v

41 CONCLUSION: Chapter 17 Quantity of Heat

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