1 / 19

Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 14 Heat. Specific Heat. Q = m c  T (c is the specific heat) Depends on the material. c = Q/(m  T) Assumes there is no phase change as the result of the heat. Q = mL (L is the latent heat of fusion or vaporization.)

umed
Télécharger la présentation

Physics 2053C – Fall 2001

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 2053C – Fall 2001 Chapter 14 Heat Dr. Larry Dennis, FSU Department of Physics

  2. Specific Heat • Q = m c T (c is the specific heat) • Depends on the material. • c = Q/(m T) • Assumes there is no phase change as the result of the heat. • Q = mL (L is the latent heat of fusion or vaporization.) • Heat required to complete a phase change (no change in temperature included).

  3. Heating water (35 gm at 30 °C) • Start at 30 °C, heat until it has all boiled away. • Step 1: heat to 100 °C (net change of 70°C. • Q = m c T • Step 2: heat change required to convert water to steam. • Q = m Lv

  4. Heating water (35 gm at 30 °C) How much heat is required to increase the temperature to 100 °C? Q = m c T Q = 35 gm * 1 cal/(gm °C ) * (70 °C ) Q = 2450 cal

  5. Heating water (35 gm at 30 °C) How much heat is required to boil the water once it reaches 100 °C? Q = m Lv Q = 35 gm * 539 cal/gm Q = 18865 cal Note it is much larger than the heat required to raise the temperature.

  6. Heating water (35 gm at 30 °C) If the water is heated at a constant rate of 100 Watts how long does it take to boil the water away? 100 W = 100 J/s 4.186 J = 1 cal So: 100 W = 100 J/s / (4.186 J/cal) 100 W = 23.9 cal/s.

  7. Heating water (35 gm at 30 °C) Step 1 (heating from 30 °C to 70 °C): Recall that Q = 2450 cal rate = 23.9 cal/s Time required = Q/rate t = 2450 cal/(23.9 cal/s) t = 102 s

  8. T t Heating water (35 gm at 30 °C) Step 2 (boiling at 100 °C): Recall that Q = 18865 cal rate = 23.9 cal/s Time required = Q/rate t = 18865 cal/(23.9 cal/s) t = 789 s

  9. Achieving Equilibrium Energy 300 g of aluminum at 80 °C is placed placed in a 150 g aluminum cup containing 400 mL of water at 27 °C. What is the final temperature of the system? EAL + EW + Ecup = 0 (mcT)AL + (mcT)W + (mcT)cup = 0 mALcAL(Tf – Ti) + mWCW(Tf-To) + mcupccup(Tf-To) = 0

  10. Achieving Equilibrium Energy 300 g of aluminum at 80 °C is placed placed in a 150 g aluminum cup containing 400 mL of water at 27 °C. What is the final temperature of the system? mALcAL(Tf – Ti) + mWCW(Tf-To) + mcupccup(Tf-To) = 0 (mALcAL+ mWCW + mcupccup)Tf = mALcALTi + (mWCW + mcupccup)To Tf = (mALcALTi + (mWCW + mcupccup)To)/(mALcAL+ mWCW + mcupccup) Tf = (.3*900*80+ (.4*4186 + .15*900)*27)/(.3*900 + .4*4186 + .15*900) Tf = 33.88 °C

  11. Achieving Equilibrium Energy 300 g of aluminum at 80 °C is placed placed in a 150 g aluminum cup containing 400 mL of water at 27 °C. What is the final temperature of the system? Energy lost by 80 °C aluminum block = mcT for the aluminum = .3 kg * 900 J/kg °C * (33.88 - 80)°C = - 12452 J Energy gained by aluminum cup = mcT for the cup = .15 kg * 900 J/kg °C * (33.88 - 27)°C = 929 J Energy gained by water in cup = mcT for the water. = .4 kg * 4186 J/kg °C * (33.88 - 27)°C = 11523 J Total change in energy = -12452 + 11523 + 929 = 0

  12. CAPA 6 A 169 g piece of lead is heated to 80 °C and then dropped into a calorimeter containing 515 g of water that initially is at 22 °C. Neglecting the heat capacity of the container, find the final equilibrium temperature of (in °C) of the lead and water. EL + EW = 0  (mcT)L + (mcT)W = 0 mLcL(Tf –TiL) + mWcW(Tf-TiW) = 0 (mLcL + mWcW)Tf= mWcWTiW + mLcLTiL Tf= (mWcWTiW + mLcLTiL )/(mLcL + mWcW)

  13. CAPA 6 A 169 g piece of lead is heated to 80 °C and then dropped into a calorimeter containing 515 g of water that initially is at 22 °C. Neglecting the heat capacity of the container, find the final equilibrium temperature of (in °C) of the lead and water. Tf= (mWcWTiW + mLcLTiL )/(mLcL + mWcW) Tf= (.169*130*80+ .515*4186*22)/(.169*130+.515*4186) Tf= 22.6 °C

  14. CAPA 8 A lead ball, with an initial temperature of 25 °C is released from a height of 120 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature of the ball after it hits. Change in Kinetic Energy + Change in Internal Energy = 0 -Mgh + McT  -Mgh + Mc(Tf – Ti) = 0 Mgh/Mc = (Tf – Ti) (Tf – Ti) = 9.8*120/(130) = 9.05 °C Tf = Ti +9. 05 °C = 34.05 °C

  15. CAPA 10 & 11: Heat Transfer by conduction. A mountain climber wears down clothing 3 cm thick with a total surface area of 1.75 m2. The temperature at the surface of the clothing is –19 °C and at the skin is 34.5 °C. Determine the rate of heat flow by conduction through the clothing assuming it is dry and that the thermal conductivity, k, is that of down. Q/t = kA(T1-T2)/L Q/t = 0.025 W/(m°C)*1.75 m2*(34.5 - -19)°C /0.03m Q/t = 78 W

  16. CAPA 10 & 11: Heat Transfer by conduction. Determine the rate of heat flow by conduction through the clothing assuming the clothing is wet, so that k is that of water and the jacket is matted down to 0.5 cm thickness. Q/t = kA(T1-T2)/L Q/t = 0.56 W/(m°C)*1.75 m2*(34.5 - -19)°C /0.005m Q/t =10486 W

  17. Quiz #9 – Heat & Energy • Specific heats, latent heats, calorimetry. • Rate of heating. • Energy changes and energy balance in heat transfer. • Note the topics that are not included: • Convection, conduction and radiation.

  18. Quiz #9 – Heat & Energy • Sample Questions. • Chap. 14: 2, 3, 5 • Sample Problems. • Chap. 14: 19, 21,25 & CAPA 2 - 7

  19. Next Time • Quiz on Chapter 14 • Begin Chapter 15 • Please see me with any questions or comments. See you on Wednesday.

More Related