**Thermodynamics**

dU = Q - W ( first law of thermodynamics )

e = 1 - T_L/T_H ( ideal engine )

dL = alpha*L_0*dT ( linear expansion )

dV = beta*V_0*dT ( volume expansion )

dQ/dt = kA(T_1 -T_2)/l ( rate of heat conduction )

R values, thermal resistance, are given by R = *l*/k.

Stefan-Boltzmann equation is dQ/dt = *e*sAT^4

dQ/dt = *e*sA(T_1^4 T_2^4) ( flow rate of heat radiation )

Q = mL ( latent heat )

T_1 is the temperature of the body, so if T_1 > T_2 then the flow of heat is body to surroundings. If T_1 < T_2 , then it is surroundings to body.

s is the constant which has value 5.67 x 10^(-8) W/m^2*K^4

e is the factor called emissivity, and it's between 0 and 1

Latent heat is the amount of energy required to change the state of a substance per mass of substance. Specific heat is the constant for a substance in a particular phase that is the amount of energy required to raise one degree of temperature of a mass of substance.

Conduction Conduction is heat transfer due to molecular collisions. This only takes place if there is a difference in temperature.

Convection - Convection is heat transfer due to mass movement of molecules from one place to another. One example is the circulation of air near a hot object. The air becomes energized and is displaced by cold air because the hot air rises.

Radiation Radiation occurs without any medium. It consists of mainly EM waves. *A good absorber is also a good emitter

The calorie is defined as the amount of heat necessary to raise the temperature of 1 gram of water by 1 Celsius degree. Heat is related to energy by 4.186J of work is equal to 1 cal. It is energy that is transferred from one body to another because of a difference in temperature. Thermal energy is the sum of all the energy of all the molecules of an object. It is equivalent to internal energy and is often used interchangeably. The mechanical equivalent of heat is 4.186J per cal.

The first law of thermodynamics states that dU = Q W.

One way to remember the sign conventions is: **ALOB of POOP**

heat **added**/**lost** is positive/opposite(negative)

work done **ON**/work done **BY** is opposite(negative)/positive.

Adiabatic means that there is no heat flow into or out of the system.

Isothermal means constant temperature.

Isochoric/isovolumetric means constant volume.

Isobaric means constant pressure.

The second law of thermodynamics has many definitions but the one that sticks best in my mind is that heat flows from hot objects to cold ones. This in turn says that natural processes tend to move toward a state of greater disorder or greater entropy.

**Problem 1
**An ideal gas has its pressure cut in half slowly, while being kept in a container with rigid walls. In the process, 265kJ of heat left the gas. a) How much work was done during this process? b) What was the change in internal energy of the gas during this process?

a) There is no work done on or by the system because of the closed container with rigid walls.

b) dU = Q W. dU = 265kJ + 0. 265kJ of heat LEFT the gas, so it is being added to the system.

**Problem 2**

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1350 J of work is done ON the gas. a) How much heat flows into or out of the gas? b) What is the change in internal energy of the gas? c) Does its temperature rise or fall?

a) Since this is an adiabatic process. Q = 0.

b) dU = Q - W. 0 (-1350J). W is -1350J because work is being done ON the gas. dU = +1350J.

c) Because dU is positive, the temperature rises.

**Problem 3**

An ideal gas expands at a constant pressure of 5 atm from 400mL to 660mL. Heat then flows out of the gas, at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate a) the total work done by the gas in the process, and b) the total heat flow into the gas.

a) W = PdV. 5 atm * 101.3*10^3N/m^2 / 1 atm * (660mL 400mL)*1m^3/10^6mL. W = 131.69J.

b) dU = 0 because the temperature reaches its original value. Q = W, Q = 131.69J.

**Problem 4**

A nuclear power plant operates at 75% of its Carnot efficiency between temperatures of 600 and 350C. If the power plant produces electric energy at the rate of 1.3GW, how much exhaust heat is discharged per hour?

e = 1-T_L/T_H . e = 1-(350+273)/(600+273), e = .286.

0.75(.286) = 1-Q L /1.3x10^9W. Q L = 2.7885x10^8W.

**Problem 5**

1.0kg of water at 30C is mixed with 1.0kg of water at 60C in a well insulated container. Estimate the net change in entropy in the system.

Q = mCdT. Q = 1kg*1kcal/kgC*(15C) = 15kCal.

-15kCal/(60+273) + 15kCal/(30+273) = 0.00446kCal/K

**Problem 6**

A 3.8kg piece of aluminum at 30C is placed in 1kg of water in a Styrofoam container at room temperature (20C). Calculate the approximate net change in entropy of the system.

-Q_lost = Q_gained. 3.8kg*0.22kcal/kgC*(30-x) = 1kg*1kCal/kgC(x-20). x = 24.55C

Q = 1kg*1kCal/kgC*(24.55-20), Q = 4.55

4.55kCal/(20+273) + -4.55kCal/(30+273) = 5.125x10^(-4)kCal/K.

**Problem 7**

a) Calculate the approximate rms speed of an amino acid whose molecular mass is 89u in a living cell at 37C. b) What would be the rms speed of a protein of molecular mass 50,000u at 37C?

a) v = sqrt(3kT/m) = sqrt(3*1.38*10^(-23)*(37+273)/(89*1.66*10^(-27)))

v = 294.7 m/s

b) v = sqrt(3*1.38*10^(-23)*(37+273)/(50000*1.66*10^(-27)))

v = 12.4 m/s

**Problem 8**

A Pyrex measuring cup was calibrated at normal room temperature. How much error will be made in a recipe calling for 0.3L of cool water, if the water and cup are hot at 80C instead of room temperature? Neglect glass expansion.

Since we neglect glass expansion, then only the water will undergo volume expansion.

dV = BV_0T. dV = 210*10^(-6)*0.3*(80). dV = 0.00504L.

0.00504/(0.00504+.3) x 100% = 1.65%.

**Problem 9
** If coal gives off 7000kCal/kg when it is burned, how much coal will be needed to heat a house that requires 4.8x10^7kCal for the whole winter? Assume that an additional 30% of the heat is lost up the chimney.

4.8*10^7kCal/( 7000kCal/kg * 0.7 ), 9795.92kg of coal

**Problem 10
**A 0.015kg lead bullet is tested by firing it into a fixed block of wood with a mass of 1.05kg. If the block and imbedded bullet together absorb all the heat energy generated and after thermal equilibrium has been reached the system has a temperature rise measured as 0.020C, estimate the entering speed of the bullet.

dT = 0.020C.

0.5*0.015kg*v^2 = (0.015kg*130J/kgC+1.05kg*1700J/kgC)*0.020C.

v = 69.03m/s

**Problem 11**

A mountain climber wears down clothing 0.035m thick with a total surface are of 1.7m^2. The temperature at the surface of the clothing is -20C and at the skin is 34C. Determine the rate of heat flow by conduction through the clothing a) assuming it is dry and that the thermal conductivity, k, is that of down, and b) assuming the clothing is wet, so that k is that of water and the jacket has matted down to 0.005m thickness.

a) dQ/dt = kA(T 1 -T 2 )/l. 0.025*1.7*(34-(-20))/0.035, 65.57J/s

b) dQ/dt = 0.56*1.7*(34-(-20))/0.005 = 10281.6J/s

** Problem 12
**Estimate the rate at which heat can be conducted from the interior of the body to the surface. Assume that the thickness of tissue is 0.04m, and that the skin is at 34C and the interior at 37C, and that the surface area is 1.5m^2. Compare this to the measured value of about 230W that must be dissipated by a person working lightly. This clearly shows the necessity of convective cooling by the blood.

dQ/dt = kA(T 1 -T 2 )/l. 0.2*1.5*(37-34)/.04 = 22.5W

**Problem 13
**At a crime scene, the forensic investigator notes that the .0082kg lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was fired at room temperature, 20C, what does the investigator calculate the minimum muzzle velocity of the gun was?

In order for the lead bullet to melt completely on impact, the bullet must have had
0.25x10^5 J/kg * .0082kg => 205J.

In addition to this energy, it must have gained enough heat to raise the temperature to 327C. q = mCdT, q = .0082kg*130J/gC*(327-20) = 327.26J.

The KE must have been at least the sum, so 532J = ½mv^2, v = 360.3 m/s

**Problem 14**

How long does it take a 750W coffeepot to bring to a boil 0.6L of water initially at 8C? Assume that the part of the pot which is heated with the water is made of .36kg of aluminum and that no water boils away.

The temperature change is going to be 100-8 = 92C. 750W means 750J/s, so we want to find the amount of energy required to boil away 0.6L of water.

Qwater = mCdT. Q = 0.6L * 1kg/L * 4186J/kgC * 92C = 231067.2J.

Qaluminum = .36kg * 900J/kgC * 92C = 29808J

(231067.2J+29808J) * 1s/750J = 347.8 seconds.

**Problem 15
**A thermosbottle has 0.6liters of water whose temperature is +12 celcius. Pieces of ice,
whose mass are a total of 0.12kg and temperature -18 celcius, are thrown into the thermos.
How much ice is going to melt? The heat capacity of the Thermosbottle is 110J/Kelvin. (Thanks Johoho)

Water's latent heat of fusion = 333kJ/kg

4.19kJ/kgC for water and 2.09kJ/kgC for ice.

Q = mCpdT

Q = mL

Q = C*dT

Final T is 0 since not all the ice melts.

Q_gained = -Q_lost

The heat from the thermos and the water goes into changing the temperature of the ice from -18C to 0C and then into the phase change from solid to liquid.

Q_rx + Q_ice = -Q_thermos - Q_water

Q_rx + Q_ice = -110J/K*1kJ/1000J*(T-12)K - (0.6l*1000cm^3/l * 1g/cm^3 * 1kg/1000g)*4.19kJ/kgC*(T-12)

0.12kg*2.09kJ/kgC*(0-(-18)) + m*333kJ/kg = -110J/K*1kJ/1000J*(0-12)K - (0.6l*1000cm^3/l * 1g/cm^3 * 1kg/1000g)*4.19kJ/kgC*(0-12)

4.5144 + m*333kJ/kg = -0.11(0-12) - 2.514(0-12)

m = 0.081kg.

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