For water near 1 atm., Lf  3.33 x 105 J/kg. Phase Diagram for Water The liquid/solid curve (A-B- ) is the melting curve. At any temperature/pressure along this curve, internal energy is needed to melt the solid, even though the temperature will not change. The latent heat of fusion is defined as Lf  Q/m, where Q is the amount of heat needed to melt a mass m of the solid (e.g. ice  water). Similarly, heat Q = m Lf flows out of the liquid when a mass m freezes. Lf will depend (weakly) on the pressure (and temperature) at which the melting or freezing occurs. For water near 1 atm., Lf  3.33 x 105 J/kg.

The curve A-C-D, which gives the vapor pressure of the liquid, is the boiling curve. Boiling occurs when the vapor pressure equals the pressure on the system, so that even bubbles of vapor can form inside the liquid. At any temperature/pressure along this curve, internal energy is needed to vaporize the liquid. The latent heat of vaporization is defined as Lv  Q/m, where Q is the amount of heat needed to vaporize a mass m of liquid. Similarly, heat Q = m Lv flows out of the vapor when a mass m condensed. Lv depends on the pressure/temperature more strongly than Lf: at the critical point (D), Lv = 0. For water at 1 atm. (100 oC), Lv = 2.26 x 106 J/kg.

at P = 1 atm = 1.01 x 105 N/m2 [Lv values are typically an order of magnitude greater than Lf values, since Lv involves breaking bonds between atoms/molecules whereas Lf involves allowing bonds to move around.]

Problem: Compare the amounts of heat that are needed to increase the temperature of 1 kg of ice from -100 oC to 0o C melt 1 kg of ice Increase the temperature of 1 kg of water from 0 oC to 100 oC boil 1 kg of water at 1 atm increase the temperature of steam from 100 oC to 200 oC at 1 atm. cice = 2090 J/kgoC; cwater = 4186 J/kgoC; csteam  2010 J/kgoC Lf = 3.33 x 105 J/kg; Lv = 2.26 x 106 J/kg Qice = mcice T = (1 kg) (2090 J/kgoC) (100 oC) = 2.09 x 105 J Qmelt = m Lf = (1kg) (3.33 x 105 J/kg) = 3.33 x 105 J Qwater = mcwater T = (1 kg) (4186 J/kgoC) (100 oC) = 4.19 x 105 J Qboil = m Lv = (1 kg) (2.26 x 106 J/kg) = 2.26 x 106 J Qsteam = mcsteam T = (1 kg) (2010 J/kgoC) (100 oC) = 2.01 x 105 J Boiling takes the most energy – breaking bonds between the molecules

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? [Assume that the mass and heat capacity of the beaker are negligible.] cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. some, but not all, of the water will freeze, so that Tf = 0 oC. Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC.

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC. What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. Not likely some, but not all, of the water will freeze, so that Tf = 0 oC. Not likely Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC. To solve, you can try to “pre-decide” which is the appropriate outcome, and try to solve assuming your guess is correct. If your guess is incorrect, you will get a “nonsense” answer (e.g. temperature higher or lower than any of the initial temperatures or a negative mass). In that case, try another possibility. If your guess is correct you will get a reasonable (i.e. non-nonsense) answer. Therefore, although “pre-deciding” may save time, it is not necessary: just try possibilities until you get an answer that makes sense.

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC. What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. some, but not all, of the water will freeze, so that Tf = 0 oC. Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC. Guess (a): all the water will freeze: Q(ice: Ti  Tf) + Q(water: Tw 0 oC) + Qfreeze + Q(new ice: 0 oC  Tf) = 0 M cice(Tf – Ti) + m cwater(0oC - Tw) – mLf + m cice (Tf - 0 oC )= 0 (M + m)cice Tf = MciceTi + mLf + mcwaterTw Tf = [(0.25)(2090)(-5) +(0.6)(3.33x105) + (0.6)(4186)(18)] / [(0.85)(2090)] oC = 136 oC: nonsense

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC. What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. some, but not all, of the water will freeze, so that Tf = 0 oC. Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC. Guess (b): Tf = 0 oC. Let the amount of water that freezes = m. Q(ice: Ti  Tf) + Q(water: Tw 0 oC) + Q (water that freezes) = 0 M cice(0 oC – Ti) + m cwater(0oC - Tw) - m Lf = 0 m = (Mcice Ti + mcwaterTw) / Lf = [(0.25) (2090) (+5) + (0.6)(4186)(-18)] / (3.33 x 105) kg m = - 0.13 kg: nonsense!

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC. What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. some, but not all, of the water will freeze, so that Tf = 0 oC. Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC. Guess (d): Tf > 0 oC. Q(ice: Ti  0 oC) + Qmelt + Q(melted ice: 0 oC  Tf) + Q (water: Tw  Tf) = 0 Mcice(0oC-Ti) + MLf + Mcwater(Tf - 0oC) + Mcwater(Tf – 0oC) + mcwater(Tf – Tw) = 0 Tf = (MciceTi – MLf +mcwaterTw) / [(M+m)cwater] Tf = [(0.25)(2090)(-5) – (0.25)(3.33x105) + (0.6)(4186)(18)] / [(0.85)(4186)] oC Tf = - 11.4 oC: nonsense

Problem: M = 0.25 kg of ice at Ti = -5 oC is added to an insulated beaker holding m = 0.60 kg of water at Tw = 18 oC. What is the final temperature, Tf? If Tf = 0 oC, how much ice is there at the end? cice = 2090 J/kgoC; cwater = 4186 J/kgoC; Lf = 3.33 x 105 J/kg Note: Before plugging in numbers, one can imagine 4 possible outcomes: all the water will freeze and the final temperature Tf < 0 oC. some, but not all, of the water will freeze, so that Tf = 0 oC. Some, but not all, of the ice will melt, so that Tf = 0 oC. all of the ice will melt and Tf > 0 oC. Guess (c): Tf = 0 oC. Let the amount of ice that melts = m. Q(ice: Ti  Tf) + Q(ice that melts) + Q(water: Tw 0 oC) = 0 M cice(0 oC – Ti) + m Lf + m cwater(0oC - Tw) = 0 m = (MciceTi + mcwaterTw)/Lf = [(0.25)(2090)(-5) + (0.60)(4186)(18)]/(3.33 x 105) kg m = 0.13 kg  So the total mass of ice at end m(ice, final) = (0.25-0.13) kg = 0.12 kg

First Law of Thermodynamics: Eint = Q + W. WORK First Law of Thermodynamics: Eint = Q + W. For a general system, the internal energy is a function of temperature, pressure, volume, magnetic field (for magnetic particles), electric field (for electric particles), …. These are all called state variables of the system. They are defined for any equilibrium state of the system. If a system changes between two equilibrium states (because of a change in one or more variables), Eint  Eint(final) – Eint(initial); i.e. Eint only depends on the initial and final states. Therefore (Q+W) only depends on the initial and final states, but Q and W may each depend on the process. For a gas, the relevant state variables (besides Eint) are P, V, and T, and these are connected by the equation of state of the gas. [For an ideal gas, the equation of state is PV = nRT.] If the volume changes by a small amount, dV, “quasi-statically” (i.e. slowly enough so that there is no turbulence and the pressure and temperature stay well defined during the change), then the work done on the gas is dW = - p dV.

Like the heat, the work depends on the process. dW = - p dV Notes: If dV < 0 (gas is compressed), dW > 0: work is done by the surroundings on the gas. If dV > 0 (gas expands), dW < 0: work is done by the gas on the surroundings. If there is a larger, quasi-static change in volume from Vi to Vf, the work done is W = -  p dV Like the heat, the work depends on the process. Vf Vi The work done on the gas for each of these compressions is the area under the curve. Although the initial and final states are equal, the work done is different for each case.

If the gas is ideal (i.e low density), p = nRT/V, so W = -nRTdV/V W = -  p dV Consider a quasi-static, “isothermal” expansion of an ideal gas. To keep the temperature constant, it is in contact with an “energy reservoir”, i.e. a material so large (i.e. heat capacity Cres Mcres  ) that heat can flow in or out without changing its temperature: T = Q/Cres  0 . Vf Vi Suppose one slowly reduces the pressure on the piston, allowing the gas to expand, while T = Ti = constant. W = -  p dV If the gas is ideal (i.e low density), p = nRT/V, so W = -nRTdV/V Since T = Ti = constant, it can come out of the integral: W -nRTidV/V W = -nRTi ln(Vf/Vi) (< 0, since Vf>Vi).

W = -nRTi ln(Vf/Vi) [= - area under curve if Vf > Vi]

Now consider an adiabatic “free expansion” of the gas Now consider an adiabatic “free expansion” of the gas. (Adiabatic means that no heat flows in or out of the system, e.g. it is in an insulated container.) Between the initial and final states, the gas rushes to fill the volume and the pressure is not defined. However no work is done and no heat flows: W = 0 and Q = 0, so Eint = Q + W = 0. For a general gas, the temperature will change (e.g. as air rushes out of a tire, it becomes colder). However, experiments (Joule) show that for a low density (i.e. ideal) gas, T = 0. Therefore, Tf = Ti for this process.

Both processes have the same initial state: Vi, Ti, Pi = nRTi/Vi and the same final state: Vf, Ti, Pf = nRTi/Vf . Therefore, both processes must the same Eint . But Eint (adiabatic free expansion) = 0. Therefore Eint (isothermal expansion) = 0. Therefore Qisothermal expansion + Wisothermal expansion = 0. Wisothermal expansion = -nRT ln (Vf/Vi)  Qisothermal expansion = + nRT ln (Vf/Vi) [Wadiabatic free expansion = Qadiabatic free expansion = 0] i.e. W and Q depend on process, whereas Eint only depends on end points.