1. A.M.NASR Chapter (3) QUANTITY OF HEAT

2. A.M.NASR Thermal energy is the energy associated with random molecular motion. It is not possible to measure the position and velocity of every molecule in a substance in order to determine its thermal energy. We can measure changes in thermal energy by relating it to change in temperature. The thermal energy lost or gained by objects is called heat. This chapter is concerned with the quantitative measurement of heat. Thermal energy

3. A.M.NASR Units of Heat Cal is the quantity of heat required to change the temperature of one gram of water through one celsius degree. Kilocalorie is the quantity of heat required to change the temperature of one kilogram of water through one celsius degree. (1 kcal =1000 cal.( British thermal unit (Btu) is the quantity of heat required to change the temperature of water with a mass of one standard pound through one Fahrenheit degree. Btu = 252 cal = 0.252 kcal

4. A.M.NASR Heat capacity The heat capacity of a body is the ratio of heat supplied to the corresponding rise in temperature of the body. Heat capacity = Q/T The specific heat capacity of a material is the quantity of heat required to raise the temperature of a unit mass through one degree. c = Q / mΔT

5. A.M.NASR Example 3.1 How much heat is required to raise the temperature of 200g of mercury from 20 to 100K? (c= 0.033 cal/g o C) Q = mcΔT = (200 g)(0.033 cal/g o C)(100 o C -20 o C) = 528 cal

6. A.M.NASR Example 3.2 During a bout with the flu an 80-kg man ran a fever of 2 o C above normal; that is, his body temperature was 39 o C instead of the normal 37 o C. Assuming that the human body is mostly water (c=1 kcal/kg o C), how much heat is required to raise his temperature by that amount?

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8. A.M.NASR Example 3.3 You are designing an electronic circuit element made of 23 mg of silicon. The electric current through it adds energy at the rate of 7.4 mW = 7.4x10 -3 J/s. If your design doesn't allow any heat transfer out of the element, at what rate does its temperature increase? The specific heat capacity of silicon is 705 J/kg.K.

9. A.M.NASR Example 3.4 A handful of copper shot is heated to 90 o C and then dropped into 80g of water at 10 o C. The final temperature of the mixture is 18 o C. What was the mass of the shot?

10. A.M.NASR Example 3.4 we have neglected two important facts:(1) the water must have a container, which will also absorb heat from the shot; (2) the entire system must be insulated from external temperatures.

11. A.M.NASR Example 3.5 In a laboratory experiment it is desired to use a calorimeter to find the specific heat of iron. Eighty grams of dry iron shot is placed in a cup and heated to temperature of 95 o C. The mass of the inner aluminum cup and of the aluminum stirrer is 60g. The calorimeter is partially filled with 150g of water at 18 o C. The hot shot is quickly poured into the cup, and the calorimeter is sealed. After the system has reached thermal equilibrium, the final temperature is 22 o C. Compute the specific heat of iron.

12. A.M.NASR Example 3.5 Q w =mc  T=(150g)(1cal/g o C)(22 o C -18 o C)= 600 cal Qal = mc  T = (60g)(0.22 cal/g o C)( 22 o C -18 o C) = 52.8 cal Heat gained = 600 cal + 52.8 cal = 652.8 cal The heat lost by the iron shot: = mc s ΔT = (80 g)c s (95C-22 o C) The heat lost equal to the heat gained (80 g)c s (73 o C) = 652.8 cal Cs =

13. A.M.NASR Change of Phase The latent heat of fusion L f of a substance is the heat per unit mass required to change the substance from the solid to the liquid phase at its melting temperature. The latent heat of vaporization L v of a substance is the heat per unit mass required to change the substance from a liquid to a vapor at its boiling temperature.

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15. Example 3.6: What quantity of heat is required to change 20 lbm of ice at 12F to steam at 212F?

16. A.M.NASR Example 3.6: What quantity of heat is required to change 20 lbm of ice at 12F to steam at 212F?

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18. Example 3.7 After 12g of crushed ice at -10 o C is dropped into a 50 g aluminum calorimeter cup containing 100g of water at 50 o C, the system is sealed and allowed to reach thermal equilibrium. What is the resulting temperature? Total heat lost = heat lost by calorimeter + heat lost by water

19. A.M.NASR Total heat gained = heat gained by ice + heat to melt ice + heat to bring water to te.

20. A.M.NASR Example 3.8 If 10g of steam at 100C is introduced into a mixture of 200g of water and 120g of ice. find the final temperature and composition of the mixture. The heat required to melt the 120g of ice completely at 0 o C. Q 1 = m i L f = (120 g)(80 cal/g) = 9600 cal The maximum heat we can expect the steam to give up is Q 2 = m s L v + m s c w (100 o C - 0 o C) = (10)(540) + (10)(1)(100) = 6400 cal To determine the final composition of the mixture, note that 3200 additional calories would be required to melt the remaining ice. Hence

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