Quantity of Heat Created by: Marlon Flores Sacedon

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Presentation transcript:

Quantity of Heat Created by: Marlon Flores Sacedon Physics section, DMPS June 2010

MFS Quantity of Heat Where: Q = quantity of heat (J) m = mass (kg) T = change in temperature (K) c = specific heat capacity (J/kg. K ) 1 cal = 4.186 J 1 kcal = 1000 cal = 4186 J 1 Btu = 778 ft.lb = 252 cal = 1055 J Specific heat capacity Specific heat capacity of water: 4190 J/kg.K 1 cal/g. Co 1 Btu/lb. Fo MFS

Example 1: During a bout with the flu an 80-kg man ran a fever of 2 Example 1: During a bout with the flu an 80-kg man ran a fever of 2.0Co above normal, that is, a body temperature of 39 oC (102.2 0F) instead of the normal 37oC (98.6 oF). Assuming that the human body is mostly water, how much heat is required to raise his temperature by that amount? Ans: 6.7x105J Example 2: You are designing an electronic circuit element made of 23 mg silicon. The electric current through it adds energy at the rate of 7.4mW = 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. Ans: 0.46 K/s

dQ = quantity of heat (Cal) n = No. of moles (mols) (Quantity of heat in terms molar heat capacity) No. of moles mass Molar mass OR (Quantity of heat in terms molar heat capacity) Molar heat capacity Specific heat capacity Where: dQ = quantity of heat (Cal) n = No. of moles (mols) dT = change in temperature (K) C = Molar heat capacity (J/mol. K ) Molar mass Molar heat capacity of water = 75.4 J/mol.K

Specific Heat Capacity, c Approximate specific and molar heat capacities (constant pressure) Substance Specific Heat Capacity, c (J/Kg.K) M (kg/mol) Molar Heat Capacity, C (J/mol.K) Aluminum 910 0.0270 24.6 Beryllium 1970 0.00901 17.7 Copper 390 0.0635 24.8 Ethanol 2428 0.0461 111.9 Ethylene glycol 2386 0.0620 148.0 Ice (near 0oC) 2100 0.0180 37.8 Iron 470 0.0559 26.3 Lead 130 0.201 26.9 Marble 879 0.0585 87.9 Mercury 138 27.7 Salt 51.4 Silver 234 0.108 25.3 Water 4190 75.4

Phase Changes Where: Q = quantity of heat (J) m mass (kg) and Where: Q = quantity of heat (J) m mass (kg) Lf = heat of fusion (J/kg) LV = heat vaporization (J/kg ) Heat of fusion of water: Lf = 3.34x105 J/ kg = 79.6 cal/g = 143 Btu/lb Heat of vaporization of water: Lf = 2.256x106 J/ kg = 539cal/g = 970 Btu/lb

Problems 1. Ans. 66 oC 2. Ans. 69 g

Problems 3. Ans. 106 oC, 0.034 g

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