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Quantity of HeatCreated by: Marlon Flores Sacedon Physics section, DMPS

June 2010

MFSMFS

TmcQ

TmQ

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 J1 kcal = 1000 cal = 4186 J1 Btu = 778 ft.lb = 252 cal = 1055 J

Specific heat capacity of water:• 4190 J/kg.K• 1 cal/g. Co • 1 Btu/lb. FodT

dQ

mc

mcdTdQ

1

Specific heat capacity

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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

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nMm

mass No. of moles

Molar mass

TnMcQthen

TmcQfrom

:

:

CMcMolar heat capacity

Specific heat capacity

Molar mass

TnCQ (Quantity of heat in terms molar heat capacity)

nCdTdQ(Quantity of heat in terms molar heat capacity)

OR

dT

dQ

nC

1

Where:dQ = quantity of heat (Cal) n = No. of moles (mols) dT = change in temperature (K) C = Molar heat capacity (J/mol. K )

Molar heat capacity of water = 75.4 J/mol.K

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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 0.201 27.7

Salt 879 0.0585 51.4

Silver 234 0.108 25.3

Water 4190 0.0180 75.4

Approximate specific and molar heat capacities (constant pressure)

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fmLQ

Phase Changes

VmLQ 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

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Problems

1.

2.

Ans. 66 oC

Ans. 69 g

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Problems

3.

Ans. 106 oC, 0.034 g