Thermodynamics is the study of heat and how it moves around. Although this gives the impression of a somewhat narrow topic, it is in fact one of the most far-reaching subjects in physics.

To illustrate this, consider a block sliding to a stop on a table. The original kinetic energy of the block is converted to heat energy. Now consider the sound and light in a room. Except for that small fraction that escapes through the doors and windows, the light and sound is converted into heat energy as it is absorbed by walls, floor and ceiling. As an ice cube melts it is merely absorbing heat out of the room which is coming to thermal equilibrium.

Heat energy, then is the end point of many (and, eventually, all) physical processes. To understand the real world in any sophisticated way, we must learn what heat is. Before one gets to the deeper philosophical implications of thermodynamics, however, it is necessary to take care of some of the nitty-gritty concepts and definitions.

Right off the bat you must understand that although heat is a form of kinetic energy molecules in motion it is random kinetic energy. Thus a flying bullet has a lot of kinetic energy, but the motion of the molecules is hardly random.

So here's the definition: Heat is the total random kinetic energy of the molecules in an object.

And next: Temperature is a measure of the average random kinetic energy of the molecules of an object.

Although heat and temperature clearly are related, they are not the same and don't have the same units. A cup of coffee is at a considerably higher temperature than the Pacific Ocean, but it has much less heat.

I have arranged a series of questions for you to answer. I think the answers will seem intuitively obvious to you, at least when you look back at them. If they aren't, take a good hard look at the answers, think about them, and think about them some more until you feel comfortable with them.

Imagine a 6 kg block of an unknown metal. 9000 joules of heat are required to raise its temperature 3°C.
a) How much heat will raise it 1°C?
b) How much will raise 1 kg of metal 3°C?
c) If you applied 9,000 joules of heat to 1 kg of the metal, what would the temperature increase be?
d) How much heat is required to raise 1 kg 1°C?

answers:
a) 3000 joule
b) 1500 joule
c) 18°C
d) 500 joule

Now try some more: If it takes 800 J to raise 1 kg of a substance 1°C,
a) How much heat energy will be required to raise 1 kg of the material 3°C?
b) ...to raise 4 kg 3°C?
c) ...to raise a mass m by a temperature T?

answers:
a) 2400 joule
b) 9600 joule
c) 800 * m * ÇT

The heat required to raise 1 kg of a material 1°C is called the ";specific heat"; of that material. Thus for the metal in the problem, the specific heat is 500 J/kg°C. Other metals, and other materials will have different specific heats. Here are some common ones:

material	specific heat
brass	390 J/kg°C
lead	130 J/kg°C
aluminum	920 J/kg°C
steel	250 J/kg°C
silver	230 J/kg°C
water	4200 J/kg°C
ice	2100 J/kg°C
steam	2100 J/kg°C
alcohol	2400 J/kg°C
iron	440 J/kg°C

note: Values are only accurate to two digits, but are stated to three (or even four) for convenience.

We may now state a mathematical relationship between heat and temperature. Since we are not now in a position to know what the temperature of absolute zero is, or even if such a thing exists, we will deal only with temperature changes.

Using the argument of the previous paragraph,

ÇQ = cmÇT

where c is the specific heat. This will be the central relationship for the chapter.

Before we start problem solving, I'd like to interject a historical note. There is another heat unit still in use. It is the calorie, and it is defined as the heat required to raise the temperature of 1 gram of water 1°C. Thus the specific heat of water in this system is 1.0 cal/gm°C, and the conversion is 1 calorie = 4.18 joule. We will not use calories, as they complicate the solution of some kinds of problems, but you would do well to make a mental note of the definition because the calorie is used frequently in the real world.

1. How much energy is required to raise 3 gm of water 5°C?

ÇQ = cmÇT ÇQ = (4200 J/kg°C)(.003 kg)(5°C) ÇQ = 63.0 J

2. If 11 gm of water loses 220 J, how much does its temperature drop?

ÇQ = cmÇT ÇT = ÇQ/cm ÇT = (220)/[(4200)(.011)] ÇT = 4.76°C

3. A 500 gm block of brass goes from 18°C to 24°C. How much heat was absorbed?

ÇQ = cmÇT ÇQ = (390)(.50)(6) ÇQ = 1170 J

4. The temperature of a small quantity of alcohol rises 3.2°C when 490 J are absorbed. What is the mass of the alcohol?

m = ÇQ/cÇT m = (490)/[(2.4 * 103)(3.2)] .0638 kg

5. What is the specific heat of glass if 8.0 gm of it drops 6.2°C when 25 J are lost?

c = ÇQ/mÇT c = (25 J)/[(.008 kg)(6.2 °C)] c = 504 J/kg°C

6. A 330 gm block of aluminum and a 450 gm block of lead are cooled from 12°C to -6°C. What is the heat lost?

ÇQ = aluminum + lead ÇQ = cmÇT + cmÇT ÇQ = (920)(.33)(18) + (130)(.45)(18) ÇQ = 6.52 * 103 J

7. 15 gm of silver sits in 6 gm of alcohol. How much energy is required to heat the whole thing from 21°C to 47°C?

ÇQ = silver + alcohol ÇQ = cmÇT + cmÇT ÇQ = (230)(.015)(26) + (2400)(.006)(26) ÇQ = 464 J

Heat Capacity is merely a measure of the heat required to raise a particular object 1°C. For example, if it takes 17 J to raise a gold ring 1°C, then the heat capacity of the ring is 17 J/°C. We will denote heat capacity with a capital C.

8. The heat capacity of a statue of Isaac Newton is 1020 J/°C. How much energy is required to raise it 7°C?

ÇQ = CÇT ÇQ = (1020 J/°C)(7°C) ÇQ = 7.14 * 103 J

9. A glass full of water is found to rise 3.6°C when 9420 J of heat energy are added to it. What is its heat capacity?

ÇQ = CÇT 9420 = C(3.6) C = 2.62 * 103 J/°C

10. A hamburger required 917 J to raise it 1.8°C. What is its heat capacity?

ÇQ = CÇT 917 = C(1.8) C = 509 J/°C

The First Law of Thermodynamics states one of the most basic principles in all of science: the law of conservation of energy. The idea is that energy, like mass, is neither created nor destroyed. It may change form, but the total energy within any closed system will remain exactly constant over any period of time. Interestingly, this important law was not discovered until the mid 19th century as a result of intensive study of how mechanical energy and heat are related. If we consider cases where other forms of energy play a negligible role, the total heat lost within a system must equal the total heat gained. Symbolically, Qlost = Qgained

11. 35 gm of lead at 27°C is cooled to a temperature T. Write an expression for the energy lost.

ÇQl = cmÇT ÇQl = (130)(.035)(27 - T) J

12. 14 gm of aluminum at 11°C is warmed to a temperature T. Write an expression for the energy gained.

ÇQg = cmÇT (920)(.014)(T - 11) J

13. Combine your solutions to 11 and 12 to find the equilibrium temperature when 35 gm of lead at 27°C is placed in contact with 14 gm of aluminum at 11°C.

ÇQl = ÇQg (130)(.035)(27 - T) = (920)(.014)(T - 11) 123 - 4.55T = 12.88T - 142 T = 15.2°C

14. A 74 gm piece of silver at 83°C is placed in 22 gm of alcohol at 19°C. What will be the final temperature when they come to equilibrium?

ÇQl = ÇQg cmÇTsilver = cmÇTalcohol (230)(.074)(83 - T) = (2400)(.022)(T - 19) 1413 - 17.0T = 52.8T - 1003 2416 = 69.8T T = 34.6°C

15. 34 gm of alcohol at 13°C are mixed with 40 gm of water at 33°C. What is the final temperature?

ÇQl = ÇQg cmÇTwater = cmÇTalcohol (4200)(.04)(33 - T) = (2400)(.034)(T - 13) 5544 - 168T = 81.6T - 1061 T = 26.4°C

16. A small transistor radio is known to have a heat capacity of 1.21 * 10⁴ J/°C. It is inadvertently dropped into a bowl of 2.3 kg of water at 13°C. If the original temperature of the radio was 28°C, what is the equilibrium temperature of the whole thing?

ÇQl = ÇQg CÇTradio = CÇTwater (1.21 * 104)(28 - T) = (4200)(2.3)(T - 13) (3.39 * 105) - (1.21 * 104T) = (9.66 * 103T)(1.26 * 105) 4.65 * 105 = 2.18 * 104T T = 21.3°C

17. 28 gm of alcohol at 19°C are poured into a glass at 14°C. If the equilibrium temperature of the mix is 16.8°C, what is the heat capacity of the glass?

ÇQl = ÇQg cmÇTalcohol = CÇTglass (2400)(.028)(19.0 - 16.8) = C(16.8 - 14.0) C = 52.8 J/°C

18. A sample of metal at 100°C is placed in a beaker of 61 gm of water at 21°C. If there is 47 gm of metal and 61 gm of water, and the final temperature is 24.2°C, what is the specific heat of the metal?

cmÇTmetal = cmÇTwater c(.047)(100 - 24.2) = (4200)(.061)(24.2 - 21.0) c = 230 J/kg°C

19. A 15 gm aluminum cup contains 24 gm of water, all at 46°C. If 37 gm of brass at 5°C is placed in the cup, what is the final temperature?

ÇQl = ÇQg cmÇTcup + cmÇTwater= cmÇTbrass (920)(.015)(46 - T) + (4200)(.024)(46 - T) = (390)(.037)(T - 5) 634.8 - 13.8T + 4637 - 100.8T = 14.4T - 72.2 5344 = 129T T = 41.4°C

Imagine a 1 gm ice cube, well below 0°C. If we begin to heat it, the temperature will rise uniformly to zero. Then an amazing thing happens... rather than get warmer, the cube will begin to melt. We will have to ut in a full 335 joules of heat energy simply to melt the cube, not raising its temperature even a fraction of a degree. Upon melting, the ice cube (by now, 1 gm of water) will continue to warm steadily until it reaches 100°C. Here it ceases to change temperature, but merely begins to boil. This time it takes a whopping 2255 joules to boil a single gram of water! Upon turning completely into steam the water again steadily rises in temperature.

Here is a graph of the proceedings:

The energy required to melt ice is called the ";heat of fusion"; and is 335 J/gm or 3.35 x 10⁵ J/kg. The energy required to evaporate water, the ";heat of vaporization";, is 2260 J/gm or 2.26 x 10⁶ J/kg.

20. How much energy is required to melt 15.3 gm of ice?

ÇQ = (qf)(m) ÇQ = (3.35 * 105 J/kg)(.0153 kg) ÇQ = 5.13 * 104 J

21. How much energy is required to evaporate 2.7 gm of water?

ÇQ = (qf)(m) ÇQ = (2.255 * 106 J/kg)(.0027 kg) ÇQ = 6.09 * 103 J

22. What energy is required to raise a 5 gm ice cube at -12°C to water at 17°C? Break this problem into 3 parts--heat the ice, melt the ice, heat the water.

ÇQ = heat ice + melt + heat water ÇQ = cmÇT + qfm + cmÇT ÇQ = (2100)(.005)(12) + (3.35 * 105)(.005) + (4200)(.005)(17) ÇQ = 2.16 * 103 J

23. How much energy is needed to raise 37 gm of water at 17°C to steam at 141°C?

ÇQ = heat water + boil + heat steam ÇQ = cmÇT + qvm + cmÇT ÇQ = (4200)(.037)(100 - 17) + (2.26 * 106)(.037) + (2100)(.037)(141 - 100) ÇQ = 9.95 * 104 J

24. A 17 gm ice cube at -11°C is placed on a 135 gm slab of aluminum at 71°C. What is the final temperature after everything comes to equilibrium? Assume all of the ice melts.

ÇQl = ÇQg cool Al = heat ice + melt + heat water cmÇT = cmÇT + qfm + cmÇT (920)(.135)(71 - T) = (2100)(.017)(11) + (3.35 * 105)(.017) + (4200)(.017)(T) 8818 - 124T = 5695 + 393 + 71.4T T = 14.0°C

25. A 2.0 kg block of silver at 70°C is placed on a large block of ice at 0°C. When equilibrium is reached, how much ice will have melted?

ÇQl = ÇQg cool silver = melted ice - m2 cm1ÇT = m2qf (230)(2.0)(70 - 0) = m2(3.35 * 105) m2 = .0961 kg

26. 3.0 gm of steam at 130°C are admitted to a 440 gm steel (c = 250 J/kg°C) container at 20°C. What is the temperature of the water when equilibrium is reached?

ÇQl = ÇQg cool steam + condense + cool water = heat steel cmÇT + qvm + cmÇT = cmÇT (2100)(.003)(30) + (2.26 * 106)(.003) + (4200)(.003)(100 - T) = (250)(.44)(T - 20) 189 + 6780 + 1260 - 12.6T = 110T - 2200 T = 85.1°C

27. 8 gm of water at 23°C are poured into a cup whose heat capacity is 197 J/°C. If the initial temperature of the cup is -7°C, how much of the water will freeze?

ÇQl = ÇQg cool water + freeze = heat cup cmÇT + qfm = CÇT (4200)(.008)(23 - 0) + (3.35 * 105)(m) = (197)(0 - (-7)) m = 1.81 * 10-3 kg

In the 18th and 19th centuries, while the science of thermodynamics was developing, it was not obvious that heat was just another form of regular old normal energy. In fact, it wasn't obvious that it was energy at all!

The link between heat energy and mechanical energy was found by James Prescott Joule himself. Joule is rumored to have spent his honeymoon measuring the temperatures of waterfalls. He found that the water was slightly warmer at the bottom of the f all than at the top, and correctly concluded that this was from a conversion of the ordered kinetic energy of the moving water just before hitting the pool into disordered heat energy when it did hit. He then went home to make quantitative measurements on the phenomenon. This was his apparatus:

As the weight fell it caused the paddles to swirl around in the water, gradually heating it. In this manner he discovered that one calorie of heat was equivalent to about 4.2 joules of mechanical energy.

From then on it was understood why rubbing two objects together causes them to heat up, and why a bullet striking a wall melts on contact. the mechanical energy in each case is converted to heat energy.

28. A 40 gm piece of lead traveling at 22 m/sec strikes a wall, absorbing all the heat of impact. How much does its temperature rise?

KE --> Heat (1/2)mv2 = cmÇT (1/2)(.04)(22)2 = (130)(.04)(ÇT) ÇT = 1.86°C

29. A 4 kg disk, 0.3 m in radius, spins at 17 rad/sec. With a specific heat of 1200 J/kg°C, how much will it heat up when it is stopped by a friction brake?

Rotational KE --> Heat (1/2)Iw2 = cmÇT (1/2)[(1/2)(4)(.3)2](17)2 = (1200)(4)(ÇT) ÇT = 5.42 * 10-3°C

30. A 10 kg lead shot drops 6.0 m onto an insulated slab. How much does the shot heat up on impact?

PEg --> Heat mgh = cmÇT gh = cÇT (9.8)(6) = (130)ÇT ÇT = .452°C

31. If Joule had 2.4 kg of water in his device (see diagram) and a 13 kg mass that drops 4.0 m, how much will the temperature of the water rise?

mgh = cmÇT (13)(9.8)(4) = (4200)(2.4)(ÇT) ÇT = .051°C

32. A 65 gm bullet, traveling at 480 m/sec, rips through a 0.4 kg can of creamed corn. If the bullet exits at 211 m/sec, and the specific heat of the corn is the same as water, how much does the corn heat up?

KEo --> KEf + Heat (1/2)m1vo2 = (1/2)m1v2 + cm2ÇT ÇT = [m1(vo2 - v2)]/(2cm2) ÇT = [(.065)(4802 - 2112)]/[(2)(4200)(.4)] ÇT = 3.60°C

33. A 25 gm steel bullet travels at 620 m/sec and strikes a wall. Assuming all the heat is absorbed by the bullet, and the bullet is initially at 20°C, what is the temperature after impact?

KE --> Heat (1/2)mv2 = cmÇT (1/2)v2 = cÇT (1/2)(620)2 = (250)(ÇT) ÇT = 769°C T = 20 + 769 T = 789°C

34. A 4 kg block of aluminum is slid along a surface with a coefficient of friction µ = 0.3. How much does the temperature of the block rise if it slides 15 m?

work = heat of friction µmgs = cmÇT µgs = cÇT (.3)(9.8)(15) = (920)ÇT ÇT = .048°C

35. A 230 gm steel spring whose stiffness constant is 2.7 * 10⁴ nt/m is compressed 0.11 m and placed in a beaker of 79 gm of alcohol. When the spring unsprings, what will be the temperature change of the mixture?

energy of spring = heating of springs + heating of water (1/2)kx2 = cmÇT + cmÇT (1/2)(2.7 * 104)(.11)2 = (250)(.23)(ÇT) + (2400)(.074)(ÇT) ÇT = 0.661°C

36. A motorized rubber belt with µ = 0.15 travels at 8.0 m/sec. If a 20 gm ice cube at 0°C is pressed against it with a force of 50 nt, what is the time required for the cube to melt?

µNs = qfm s = vt µNvt = qfm (.15)(50)(8t) = (335 * 103)(.12) t = 670 sec

37. A 1.2 kg rubber ball rolls off a 3.0 m ledge and bounces back to a 2.2 m height. If the specific heat of rubber is 1100 J/kg°C, what is the change in temperature? Assume that all the heat of impact goes into the ball.

ÇPE --> ÇHeat mgÇh = cmÇT gÇh = cÇT (9.8)(0.8) = (1.1 * 103)(ÇT) ÇT = 7.13 * 10-3°C

38. A can of beans has a heat capacity of 3300 J/°C. A 17 gm bullet, traveling at 820 m/sec is slowed to 290 m/sec when it goes through the can. Assuming all heat generated goes into the beans, what is the change in temperature of the beans?

KEbullet --> Qcan + KEbullet (1/2)mvo2 = CÇT + (1/2)mv2 (1/2)(.017)(820)2 = (3300)(ÇT) + (1/2)(.017)(290)2 ÇT = 1.51°C

39. An 77 kg astronaut, freely floating at 6 m/sec is hit by a large 36 kg lemon cream pie moving oppositely at 9 m/sec. How much heat energy is generated by the collision?

ÇQ = ÇKE To get final KE we must use momentum conservation to get the final v. Thus: m1v1 + m2v2 = (m1 + m2)v (36)(9) + (77)(-6) = (36 + 77)v v = -1.2 m/sec ÇKE = [(1/2)(36)(9)2 + (1/2)(77)(6)2] - (1/2)(36 + 77)(-1.2)2 ÇKE = 2760 J heat generated = KE lost. heat generated = 2760 J

40. To conserve energy, Marco DeStinkshun constructs a very tall shower. Although cold water is released at 20°C, upon striking Marco it heats to 46°C. What is the height of the shower head?

PEg --> Heat mgh = cmÇT gh = cÇT (9.8)h = (4200)(46 - 20) h = 1.11 * 104 m note: this is about 7 miles

41. A 2.2 kg meteor falls to Earth from deep space. If its specific heat is 310 J/kg°C, how much will its temperature rise upon impact?

G(Mm/R) = cmÇT [(2/3) * 10-10][(5.98 * 1024)/(6.37 * 106)] = (310)(ÇT) ÇT = 2.02 * 105°C note: this is why large meteors liquefy themselves and their surroundings on impact