Section 4.4: Heat Capacity & Specific Heat

1. Section 4.4: Heat Capacity & Specific Heat

2. The Heat Capacity of a substance is defined as: Cy(T)  (đQ/dT)y • Subscript y says that property y of the substance is held constant when Cy is measured.

3. The Heat Capacity of a substance is defined as: Cy(T)  (đQ/dT)y • Subscript y says that property y of the substance is held constant when Cy is measured. • When using this definition, recall that (đQ/dT)yis NOT a derivative because đQisn’t an Exact Differential! • Instead, (đQ/dT)yis a ratio of the 2 infinitesimal quantities đQ & dTevaluated at constant y.

4. Heat Capacity: Cy(T)  (đQ/dT)y • Specific Heat per Kilogram of mass m: mcy(T)  (đQ/dT)y • Specific Heat per Mole of υ moles: υcy(T)  (đQ/dT)y

5. Heat Capacity • The Heat Capacity is • different for each substance. • It also depends on temperature, • volume & other, parameters. • If substance A has a higher • heat capacity than substance B, it means that • More Heat is Needed • to cause A to have a certain temperature rise • ΔT than is needed to cause B to have the • same rise in temperature.

6. Some Specific Heat Values

7. More Specific Heat Values

8. More Specific Heat Values

9. Infinitesimal, Quasistatic Processes • 1st Law of Thermodynamics: • đQ = dĒ + đW(1)

10. Infinitesimal, Quasistatic Processes • 1st Law of Thermodynamics: • đQ = dĒ + đW(1) • 2nd Law of Thermodynamics: • đQ = TdS(heat reservoir) (2) • dS = Entropy Change

11. Infinitesimal, Quasistatic Processes • 1st Law of Thermodynamics: • đQ = dĒ + đW(1) • 2nd Law of Thermodynamics: • đQ = TdS(heat reservoir) (2) • dS = Entropy Change • S is a state function, so dSis an exact differential. Combining (1) & (2) gives: • TdS = dĒ + đW

12. The combined1st & 2nd Laws for infinitesimal quasistatic processes: • đQ = TdS = dĒ + đW (3) • (We’ll use this form repeatedlyin Ch. 5!)

13. The combined1st & 2nd Laws for infinitesimal quasistatic processes: • đQ = TdS = dĒ + đW (3) • (We’ll use this form repeatedlyin Ch. 5!) • Use this result with the definition of Heat Capacityat constant parameter y: • Cy(T)  (đQ/dT)y (4)

14. The combined1st & 2nd Laws for infinitesimal quasistatic processes: • đQ = TdS = dĒ + đW (3) • (We’ll use this form repeatedlyin Ch. 5!) • Use this result with the definition of Heat Capacityat constant parameter y: • Cy(T)  (đQ/dT)y (4) • For parameter y, entropy S = S(T,y). So formally, we can write the exact differential of S as • dS = [(S/T)y]dT + [(S/y)T]dy (5)

15. Combining (3) and (5) we can write: • đQ = TdS • = T[(S/T)y]dT + T[(S/y)T]dy

16. Combining (3) and (5) we can write: • đQ = TdS • = T[(S/T)y]dT + T[(S/y)T]dy • Use this result with the definition of Heat Capacityat constant parameter y: • Cy(T)  (đQ/dT)y

17. Combining (3) and (5) we can write: • đQ = TdS • = T[(S/T)y]dT + T[(S/y)T]dy • Use this result with the definition of Heat Capacityat constant parameter y: • Cy(T)  (đQ/dT)y • This gives the GENERAL RESULT: • Cy(T)  T(S/T)y

18. 1st Law of Thermo: đQ = dĒ + đW (a) • If the volume V is the only external parameter, • đW = pdV.

19. 1st Law of Thermo: đQ = dĒ + đW (a) • If the volume V is the only external parameter, • đW = pdV. • In constant volume conditions (dV = 0) & • đQ = dĒ • So, the Heat Capacity at Constant Volume is: • CV(T)  (đQ/dT)V = (Ē/T)V

20. 1st Law of Thermo: đQ = dĒ + đW (a) • If the volume V is the only external parameter, • đW = pdV. • In constant volume conditions (dV = 0) & • đQ = dĒ • So, the Heat Capacity at Constant Volume is: • CV(T)  (đQ/dT)V = (Ē/T)V • However, if the Pressure p is held constant, the • 1st Law must be used in the form đQ = dĒ + đW • So, the Heat Capacity at Constant Pressure • has the form: Cp(T)  (đQ/dT)p

21. 1st Law of Thermo: • đQ = dĒ + đW = dĒ + pdV

22. 1st Law of Thermo: • đQ = dĒ + đW = dĒ + pdV • Heat Capacity • (Constant Volume): • CV(T)  (đQ/dT)V = (Ē/T)V

23. 1st Law of Thermo: • đQ = dĒ + đW = dĒ + pdV • Heat Capacity • (Constant Volume): • CV(T)  (đQ/dT)V = (Ē/T)V • Heat Capacity • (Constant Pressure): • Cp(T)  (đQ/dT)p

24. Clearly, in general, Cp ≠ CV. • Further, in general, Cp > CV. • However, the measured Cp& CV • are very similar for solids & • liquids, but very different for • gases, so be sure you know • which one you’re using if you • look one up in a table!

25. Heat Capacity Measurements for Constant Volume Processes (cv) Insulation Insulation DT • Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal energy, Ē. The 1st Law is(if no work is done!): Q = Ē2 - Ē1 = DĒ = mcvDT Heat Q added m m

26. Dx m m Heat Capacity for Constant Pressure Processes (cp) DT • Heat is added to a substance of mass m held at a fixed pressure, which causes a change in internal energy, Ē, ANDsome work pV. So, The 1st Law is: Q = DĒ + W = mcpDT Heat Q added

27. Experimental Heat Capacity • Experimentally, it is generally • easier to add heat at constant • pressure than at constant volume. • So, tables typically report Cp for • various materials.

28. Calorimetry Example:Similar to Reif, pages 141-142 • A technique to Measure Specific Heat is to heat a sample of material, add it to water, & record the final temperature. • This technique is called Calorimetry. Calorimeter A device in which heat transfer takes place.

29. Typical Calorimeter Calorimetry. • Calorimeter: A device in which heat transfer takes place. • A typical calorimeter is shown in the figure. Conservation of Energy requires that the heat energy Qs leaving the sample equals the heat energy that enters the water, Qw. This gives: Qs + Qw = 0

30. Qs + Qw= 0 (1) • Sample Properties: Mass = ms.Initial Temperature= Ts. Specific Heat = cs(unknown)

31. Qs + Qw= 0 (1) • Sample Properties: Mass = ms.Initial Temperature= Ts. Specific Heat = cs(unknown) • Water Properties: Mass = mw.Initial Temperature= Tw. Specific Heat = cw(4,286 J/(kg K))

32. Qs + Qw= 0 (1) • Sample Properties: Mass = ms.Initial Temperature= Ts. Specific Heat = cs(unknown) • Water Properties: Mass = mw.Initial Temperature= Tw. Specific Heat = cw(4,286 J/(kg K)) • Final Temperature (sample + water) = Tf

33. Put all of this into Equation (1) Qs + Qw = 0 (1)

34. Put all of this into Equation (1) Qs + Qw= 0 (1) • Assume cW& csare independent of temperature T. Use Qs mscs(Tf– Ts ) Qwmwcw(Tf– Tw)

35. Put all of this into Equation (1) Qs + Qw= 0 (1) • Assume cW& csare independent of temperature T. Use Qs mscs(Tf– Ts ) Qwmwcw(Tf– Tw) • This gives: mscs(Tf– Ts ) + mwcw(Tf– Tw) = 0

36. Qs + Qw = 0 (1) mscs(Tf – Ts ) + mwcw(Tf – Tw) = 0 • Solve for cs & get: • Technically, the mass of the container should be included, but if mw >> mcontainer it can be neglected.

37. Consider now a slightly different problem. 2 substances A & B, initially at different temperatures TA & TB. • The specific heatscA& cB are known. • The final temperature Tfis unknown.All steps are the same as the previous example until near the end! QA + QB = 0(1)

38. QA + QB = 0 (1) • Sample Properties: Mass A = mA.Initial Temperature= TA. Specific Heat = cA Mass B = mB.Initial Temperature= TB. Specific Heat = cB • Final Temperature (A + B) = Tf (Unknown)

39. QA + QB = 0 (1) • Assume:cA& cBare independent of temperature T. • Put QA mAcA(Tf– TA) & QB mBcB(Tf– TB) into(1): mAcA(Tf– TA ) + mBcB(Tf– TB) = 0 • Solve for Tf & get: Tf = (mAcATA + mBcBTB)∕(mAcA + mBcB)