Physics 2053C – Fall 2001

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.