1. Prof. Marlon Flores Sacedon
Quantity of Heat
Prof. Marlon Flores Sacedon
Department of Mathematics and Physics
College of Arts and Sciences
Visayas State University, Visca
Baybay City, Leyte, Phiippines

2. Quantity of Heat
When you put a cold spoon into a cup of hot coffee, the spoon warms up and the coffee cools down as they approach thermal equilibrium.
The interaction that causes these temperature changes is fundamentally a transfer of energy from one substance to another.
Energy transfer that takes place solely because of a temperature difference is called heat flow or heat transfer, and energy transferred in this way is called Quantity of Heat (Q).

3. 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

4. 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

5. 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

6. 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: An aluminum tea kettle with mass 1.50 kg and containing 1.80 kg of water is placed on a stove. If no heat is lost to the surface surroundings, how much heat be added to raise the temperature from 20oC to 85oC
Ans: 5.79X105J

7. 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

8. Problem
Cup
Coffee
70 𝒐 𝑪
1) A geologist working in the field drinks her morning coffee out of an aluminum cup. The cup has a mass of 0.120kg and its initially at 20 𝒐 𝑪 when she pours in 0.300kg of coffee initially at 70 𝒐 𝑪 . What is the final temperature after the coffee and the cup attain thermal equilibrium? (Assume that coffee has the same specific heat as water and that there is no heat exchange with the surroundings.)
Thermal
Equilibrium
𝑇 𝑓𝑖𝑛𝑎𝑙
20 𝒐 𝑪
𝟎.𝟏𝟐𝟎𝒌𝒈
𝑄 𝑐𝑢𝑝 + 𝑄 𝑐𝑜𝑓𝑓𝑒𝑒 =0
𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 𝑇 𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑢𝑝 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 𝑇 𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑓𝑓𝑒𝑒 =0
𝑇 𝑓𝑖𝑛𝑎𝑙 = 𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 𝑇 𝑐𝑢𝑝 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 𝑇 𝑐𝑜𝑓𝑓𝑒𝑒 𝑚 𝑐𝑢𝑝 𝑐 𝐴𝑙 + 𝑚 𝑐𝑜𝑓𝑓𝑒𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟
=66 𝒐 𝑪

9. Phase Changes
Plasma
FORMS OF
SUBSTANCE

10. Phase Changes
Freezing
Freezing - the substance changes from a liquid to a solid

11. Phase Changes Melting- the substance changes from a solid to a liquid
Melting (Fusion)
Freezing
Melting- the substance changes from a solid to a liquid

12. Phase Changes
Condensation
Melting (Fusion)
Freezing
Condensation- the substance changes from a gas to a liquid

13. Phase Changes
Vaporization
Condensation
Melting (Fusion)
Freezing
Vaporization- the substance changes from a liquid to a gas

14. Phase Changes
Vaporization
Condensation
Sublimation
Melting (Fusion)
Freezing
Sublimation- the substance changes directly from a solid to a gas without going through the liquid phase

15. Phase Changes
Vaporization
Condensation
Sublimation
Deposition
Melting (Fusion)
Freezing
Deposition- the substance changes directly from a gas to a solid without going through the liquid phase

16. 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

18. Problem
2) A physics student wants to cool 0.25 kg of Diet Omni-Cola (mostly water), initially at 25 𝒐 𝑪 , by adding ice initially at −20 𝒐 𝑪 . How much ice should she add so that the final temperature will be 0 𝒐 𝑪 with all ice melted if the heat capacity of the container may be neglected?
Ans: 69 grams
Apply conservation of energy to the system (Cola and ice)

19. Problem
3) A heavy copper pot of mass 2.0 𝑘𝑔 (including the copper lid) is at a temperature of 150 𝒐 𝑪 . You pour 0.10𝑘𝑔 of water at 25 𝒐 𝑪 into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.
a. Final temperature
b. Phase change of water

20. Assignment

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27. Assignment
17.57

28. eNd