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

2. 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 = 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

3. 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 ( F) 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

4. 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

5. 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
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

6. 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

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

9. Problems 3.
Ans. 106 oC, g