# The First Law of Thermodynamics

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1 The First aw of Thermodynamics Q and W are process (path)-dependent. (Q W) = E int is independent of the process. E int = E int,f E int,i = Q W (first law) Q: + heat into the system; heat lost from the system W: + work done by the system. work done on the system First law of thermodynamics is an extension of the principle of energy conservation to systems that are not isolated. Problem The figure here shows four paths on a p-v diagram which a gas can be taken from state i to state f. Rank the paths according to the following parameters, greatest first. A) the change E int All paths start at i and end at f, therefore all paths have the same change in internal energy, E int

2 Problem The figure here shows four paths on a p-v diagram which a gas can be taken from state i to state f. Rank the paths according to the following parameters, greatest first. B) the greatest work W done by the gas W = dw = V V i f pdv Path 4 has the maximum area under the p-v curve. Problem The figure here shows four paths on a p-v diagram which a gas can be taken from state i to state f. Rank the paths according to the following parameters, greatest first. C) the magnitude of the energy transferred as heat Q. E int = Q W => Q = E int + W

3 Problem The figure here shows four paths on a p-v diagram which a gas can be taken from state i to state f. Rank the paths according to the following parameters, greatest first. A) the change E int All paths start at i and end at f, therefore all paths have the same change in internal energy, E int B) the greatest work W done by the gas W = dw = Vf Vi pdv Path 4 has the maximum area under the p-v curve. C) the magnitude of the energy transferred as heat Q. Q = E int + W W is greatest for path 4 => Q 4 is greatest. Some special cases of the First aw of Thermodynamics Adiabatic processes: system insulated, no heat transfer Q = 0 therefore E int = Q W = W Constant-volume process: V is fixed dw = pdv = 0, W = 0 therefore E int = Q W = Q Cyclic processes: System goes back to the initial state E int = 0 therefore Q = W

4 Some special cases of the First aw of Thermodynamics Free expansion: insulated => no heat transfer => Q = 0, expands into vacuum without pushing something to change its volume => W = 0, Therefore: E int = Q W = 0 0 = 0 et 1.00 kg of liquid water at 100 o C be converted to steam at 100 o C by boiling at standard atmospheric pressure (1 atm) as shown. The volume of that water changes from an initial value of 1.00 x10-3 m 3 as a liquid to m 3 as steam. A) How much work is done by the system during this process? W = dw = V V i f pdv = (1.01x10 5 )( )J = kj B) How much energy is transferred as heat during the process? Q = m V = (1.00)(2256) = 2256 kj C) What is the change in the system s internal energy during the process? E int = Q W = 2256 kj kj = 2087 kj

5 Daily Quiz, November 17, grams of ice at 20 o C is placed in a glass containing 100 grams of liquid water at 10 o C. Assume that there is no heat transfer from the water to the glass or to the environment. What is the state when the ice/water mixture has come to thermal equilibrium? F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid water at 0 o C Daily Quiz, November 17, 2004 First: find how much energy is available in cooling the water to 0 o C. Q water = m water c water ( T water ) = (100g)(4.190 J/g o C)(10 o C) = 4,190 J 50 grams of ice at -20 o C is placed in a glass containing 100 grams of liquid water at 10 o C. Assume that there is no heat transfer from the water to the glass or the environment. What is the state when the ice/water mixture has come to thermal equilibrium? F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid at 0 o C

6 Daily Quiz, November 17, 2004 Second: find how much energy is needed to warm the ice to 0 o C. Q ice = m ice c ice ( T ice ) = (50g)(2.22 J/g o C)(20 o C) = 2,220 J 50 grams of ice at -20 o C is placed in a glass containing 100 grams of liquid water at 10 o C. Assume that there is no heat transfer from the water to the glass or the environment. What is the state when the ice/water mixture has come to thermal equilibrium? F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid at 0 o C Daily Quiz, November 17, 2004 Third: find how much energy is needed to melt all of the ice. Q melt = m ice F = (50g)(333 J/g) = 16,650 J 50 grams of ice at -20 o C is placed in a glass containing 100 grams of liquid water at 10 o C. Assume that there is no heat transfer from the water to the glass or the environment. What is the state when the ice/water mixture has come to thermal equilibrium? F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid at 0 o C

7 Daily Quiz, November 17, 2004 Clearly, enough energy is available to raise the temperature of the ice (Q water = 4,190J) > (Q ice = 2,220J) to 0 o C... but not enough to melt all of the ice (Q melt = 16,650J). 50 grams of ice at -20 o C is placed in a glass containing 100 grams of liquid water at 10 o C. Assume that there is no heat transfer from the water to the glass or the environment. What is the state when the ice/water mixture has come to thermal equilibrium? F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid at 0 o C Daily Quiz, November 17, 2004 How much ice melted? (Q water = 4,190J) (Q ice = 2,220J) = 1,970J available to melt ice. Q melt = m melt F => m melt = Q melt / F = (1970J)/(333J/g) = 5.9g Final state: 44.1grams of ice at 0 o C and grams of liquid water at 0 o C. F = 333 J/g c ice = 2.22 J/g o C c water = J/g o C 3) some ice, some liquid at 0 o C

8 Heat Transfer Mechanisms There are three heat transfer mechanisms: Conduction: heat transfer through direct contact. Convection: Convection happens when a fluid comes in contact with an object whose temperature is higher than that of the fluid. Heat is transferred through the flow of the fluid. Radiation: Heat can be exchanged via electromagnetic waves, called thermal radiation. It does not need a medium. Heat Transfer Mechanisms Conduction: heat transfer through direct contact. T Q = ka T H k: thermal conductivity, different for different material. The higher the k value, the better a thermal conductor it is. t e.g. Aluminum: k = 235; window glass: k = 1 (unit: W/m. K) Conduction rate: P cond = Q/t = ka (T H T ) /

9 Thermal resistance of a slab of thickness : R = / k The higher the R-value, the better a thermal insulator it is. e.g. R-13 fiberglass insulation for the house: R = 13 ft 2 o F. h/btu How much heat is lost through 1 ft 2 of this fiberglass in 24 hr period when T outside = 32 o F and T inside = 72 o F? T Q = ka H T TH T t = A R t o o 2 72 F 32 F = (1ft ) (24h) = 73.8 Btu 2 o 13ft Fh / Btu Between R-11 and R-13 fiberglass material for insulation, which one will you pick for your house? The energy transferred through one material in a certain time must be equal to that transferred through the other material in the same time. i.e. P 1,cond = P 2,cond P cond k 2A(TH TX ) k1a(tx TC ) = = solve for T X : Conduction through a composite Slab 2 P cond A(TH TC ) = / k + / k For n layers: P cond = n A(T T ) i= 1 H i C ( / K ) i examples: dry wall + insulation +outside wall for your house, and shirt + sweater + coat for your body in the winter.

10 Convection: Convection happens when a fluid comes in contact with an object whose temperature is higher than that of the fluid. Heat is transferred through the flow of the fluid. Think of warm air rising from a heat register Radiation: Heat can be exchanged via electromagnetic waves, called thermal radiation. It does not need a medium. P rad = σεat 4 σ = x10-8 W/m 2. K 4 (Stefan-Boltzmann constant) ε: emissivity, a value between 0 and 1 A: surface area T: temperature of the surface in Kelvin (Classical Approximation -- corrected by Quantum Mechanics)

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