Thermodynamics Heat and Work Work and energy Recall

Thermodynamics: Heat and Work

Work and energy • • Recall W = ΔKE Work can increase the internal energy of a substance Internal energy decreases by giving off heat. The reverse can also happen

First Law of Thermo • ΔU = Q – W, • The change in internal energy (ΔU) equals the heat (Q) added to the system minus the work (W) done by the system.

First Law Explained • Remember that heat is kinetic energy and work is a change in kinetic energy. • To do work ON something means to GIVE it energy (as either heat or motion) • To do work means to SPEND energy.

Consider this • What is meant by internal energy again? • Can internal energy be negative? • What does this mean for the first law?

Question 1 • What happens to the internal energy of water vapor in the air that condenses on the outside of a cold glass of water? Is work done or is heat exchanged?

Answer • It decreases • Heat transfer – Water condensating on a glass doesn’t move it or break it (no work)

Math Example • 2500 J of heat is added to a system, and 1800 J of work is done ON the system. What is the change in internal energy of the system?

Solution • Use ΔU = Q – W, • ΔU = 2500 J – (-1800 J) = 4300 J • NOTE: work is negative because it is being done ON the system.

Math Example 2 • 2500 J of heat is added to a system and 1800 J of work is done BY the system. What is the change in internal energy of the system?

Solution • Again use ΔU = Q – W, • ΔU = 2500 J – 1800 J = 700 J • NOTE: work is positive here because work is done BY the system. • On = negative • By = positive

The system • So far we have treated the internal energy of a substance or a combination of substances as a single quantity to or from which energy is transferred. • Such a substance or combination is called a system.

Examples of Systems • • The hammer, nail, and board The water and metal A hot air balloon A car engine

In real life • Systems are almost never completely isolated from its surroundings. • A system interacts with its environment.

Work done on gasses • Again recall, Work = force x distance • Recall pressure = force / area and volume = area x distance. • These three facts lead us to the following conclusion: W = Fd = PΔV = pressure x change in volume

Important! • W = PΔV makes one very important assumption. Can you guess what that assumption is? • That temperature remains constant.

Example • A cylindrical piston with an area of 0. 025 m 2 is pushed 0. 05 m by a gas that exerts a constant pressure of 9 x 105 Pa. How much work was done by the gas on the piston?

Solution • • A = 0. 025 m 2 d = 0. 05 m P = 9 x 105 Pa = 9 x 105 N/m 2 W = PΔV = PAd = 9 x 105 N/m 2(0. 025 m 2)(0. 05 m) = 1125 J

Question 2 • Use the conservation of energy to explain why the temperature of a gas increases when it is compressed and decreases when it expands.

Answer • Compressing a gas takes work. Doing work on the gas increases its internal energy in the form of heat. • Expanding gasses are doing work (either pushing other gasses out of the way or pushing the walls of its container out) and losing energy.

Thermodynamic Processes • In reality energy is transferred as both heat and work to some extent. • However, in most cases one type of energy transfer dwarfs the other. • Therefore, we can use ideal processes to approximate real life ones.

On strike • When a gas changes temperature but does not change in volume no work is done. • Such a process is called a constant volume process or isovolumetric process • Iso = same • Volumetric = volume

Conceptual Example • A bomb calorimeter is an enclosure in which substances undergo a combustion reaction. • The gas released by the reaction has nowhere to go. All the energy is transferred as heat. • This device tells you how much energy is released by the reaction.

Staying cool • When a gas expands slowly enough it stays in thermodynamic equilibrium with the environment. • As it expands it does work on the environment but no net heat is transferred. • Such a process is called isothermal • Iso = same • Thermal = temperature

Conceptual Example • Picture a partially inflated balloon. • If the atmospheric pressure decreases, like before a storm, the gas will expand. • As it expands it decreases in internal energy and temperature. • However, heat transfers from the air into the balloon, keeping it a constant temperature.

Question 3 • In an isothermal process, 3700 J of work is done by an ideal gas. Is this enough information to tell how much heat has been added to the system? If so, how much?

Answer • In an isothermal process there is NO net heat transfer. • So, we do have enough info. • There is 0 heat added.

Question 4 • Is it possible for the temperature of a system to remain constant even though heat flows into or out of it? • Yes, as long as the system is free to expand or contract.

Rapid Expansion • Isothermal processes must happen slowly. • If a gas expands rapidly its temperature, pressure, and internal energy decrease. • If this happens in a closed environment, no heat can be transferred to or from the environment, such a process is called an adiabatic process from a Greek word meaning impassible

Question 5 • Explain why the temperature of a gas increases when it is adiabatically compressed.

Answer • Adiabatically means the heat can’t get out. • The trapped heat must be used to increase the internal energy of the gas, which will increase its temperature.

One Way Street • In nature many things happen spontaneously while the reverse thing does not. • Heat flow from hot to cold • Falling objects creating heat • A falling glass breaking

nd 2 law of thermodynamics • There are several ways to state this law but here is the most common: “Heat flows naturally from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object” – R. J. E. Clausius

nd 2 law and heat engines • Heat engine – any device that changes thermal energy into mechanical work. • Examples: steam engines and car engines • Work can only be obtained when heat is allowed to flow from a high temperature to a low temperature

How it works • A heat input, QH, at a high temperature, TH, is partly transformed into work, W, and partly exhausted as heat, QL, at a lower temperature, TL

Operating Temperature • TH and TL are called the operating temperatures of the engine and are important in calculating how efficient your engine is.

Efficiency • The efficiency, e, of any heat engine is the ratio comparing the work it does to the heat input it uses. • e = W/QH • Conservation of energy tells us: QH = W + QL or W = QH – QL • This changes e = W/QH to (QH – QL) / QL or 1 – (QL / QH)

Example • An automobile engine has an efficiency of 20 percent and produces an average of 23, 000 J of mechanical work per second. How much heat is discharged from this engine per second?

Solution • • W = 23, 000 J e = 0. 20 e = 1 – (QL / QH) QL / QH = 1 – e = 0. 80 Also QH = W/e = 23, 000 J / 0. 20 = 1. 15 E 5 J QL = 0. 80 QL = (0. 80)(1. 15 E 5 J) = 9. 2 E 4 J This is 92, 000 watts.

Efficiency in terms of Temp Because heat is proportional to temperature we can rewrite our efficiency equation: eideal = (TH – TL) / TH = 1 – (TL / TH) We call this e the eideal because this would be the efficiency of a perfect engine, one that takes in heat and uses all of it for work.

D. N. E • This leads us to our next statement of the 2 nd law: • “No device is possible whose sole effect is to transform a given amount of heat completely into work. ” – Kelvin – Planck • Translation: a perfect engine does not exist

Question 6 • Is it possible to cool down a room on a hot summer day by leaving the refrigerator door open?

Answer • No way, the refrigerator must give off waste heat, which would be released into the room.

Question 7 • Which will give the greater improvement in the efficiency of a Carnot engine, a 10 o. C increase in the high-temperature reservoir, or a 10 o. C decrease in the low-temperature reservoir?

Answer • • • Start with the fraction 3/4. What is bigger, 2/4 or 3/5? 3/5 obviously, Because e = 1 – TL / TH, 1 – 2/4 is greater than 1 – 3/5. Dropping the lower temp is better.

In other words • The previous statement (the Kelvin – Planck statement) is commonly explained this way: • You can’t get something for nothing (1 st law) • You can’t even break even (2 nd law)

Entropy • Entropy: a measure of the order or disorder of a system • When dealing with entropy it is the change in entropy that we care about. • ΔS = Q / T • Where S is entropy, Q is still heat, and T is the temperature in Kelvin

Question 8 • The oceans contain a tremendous amount of thermal energy. Why, in general, is it not possible to put this energy to useful work?

Answer • The ocean also has a very large entropy, every water molecule is going every which way. • Imagine a bunch of 5 year olds running around in all directions, it’s a lot of energy but you couldn’t use it to push a boulder unless they were all going the same direction.

Example • An ice cube of mass 60 g is taken from a storage compartment at 0 o. C and placed in a paper cup. After a few minutes, exactly half of the mass of the ice cube has melted, becoming water at 0 o. C. Find the change in entropy of the ice/water.

Solution • Step 1: find Q – The heat needed to melt 30 g of ice is found using the latent heat of fusion. – Q = m. L = (30 g)(79. 7 cal/g) = 2400 cal = 2. 4 Cal • Step 2: Find ΔS – ΔS = Q / T = 2400 cal / 273 K = +8. 8 cal / K

But wait • In the last example the temperature stayed the same. • What happens to the entropy (or disorder) of the system if the temperature changes? • Answer: ΔStotal = ΔSH + ΔSC, where ΔSH is the entropy at the higher temp and ΔSC is the entropy at the lower temp.

Example • Find the change in entropy when 50. 0 kg of water at 20. 0 o. C is mixed with 50. 0 kg of water at 24. 0 o. C.

Solution • Step 1: find Q – Q = mc ΔT = (50. 0 kg)(1. 00 kcal /kg*K)(2. 0 o. C) = 100 kcal – Because equal masses of water were mixed the final temperature will be the average of the 2, or 22 o. C

Solution • Step 2: find ΔSH = -100 kcal / 296 K = -0. 338 kcal / K – Note Q is negative because the hot water gives off heat. • Step 3: find ΔSC = 100 kcal / 294 K = 0. 340 kcal • Step 4: add, ΔS = -0. 338 + 0. 340 = 0. 002 kcal/K

Notes • Notice in the last example that even though the entropy went down for one part of the system, the entropy for the total system when up. • ΔS > 0 • If the system is not isolated, then the total entropy is the entropy of the system plus the entropy of the environment. • ΔS = ΔSs + ΔSenv 0

nd Entropy and the 2 Law • We have finally come to our general statement of the 2 nd Law: • The total entropy of any system plus that of its environment increases as a result of any natural process. • This can also be stated as: Natural processes tend to move toward a state of greater disorder.

Information Theory • It is accepted that a more orderly system requires more information to describe it and a less orderly system requires less information.

Conceptual example • • Remember our falling rocks creating heat when it hits the ground? Why can’t rocks absorb heat and jump up into the air? The random motion of thermal energy is 1 piece of info. In a falling body all molecules fall at the same speed as well as randomly due to thermal energy. (2 pieces)

Question 9 • What happens if you open a container of chlorine gas? Does the reverse ever happen?

Answer • The gas molecules will spread out and fill the room. • No, the molecules will not randomly all gather back into the bottle.

Question 10 • Suppose you collect a lot of papers scattered all over the floor and put them in a neat stack; does this violate the second law of thermodynamics? Explain.

Answer • The work you did organizing the papers disturbed the air in the room and created more heat, both of which will increase the entropy of the system MORE than the decrease in the entropy of the papers.

Heat Death • As a system goes from orderly to disorderly the amount of work you can get out of the system decreases. • Well since the entropy of the universe is always increasing, the logical conclusion is that eventually the universe will be a uniform mixture at thermal equilibrium. • No work can happen and no change will occur, this is called the Heat Death of the Universe (note: assumes a finite universe)