Heat Transfer Laws
Kirchhoff’s law according to this aw, the ratio of emissive power to adsorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. Thus for a body having e and a as emissive and absorptive power
e/a = E/ A = E/1 = E
Where E = emissive power of a black body
It implies a good absorber and is also a good emitter (radiator)
Fraunhofer lines they are the dark lines in the spectrum of the sun and are explained on the basis of kirchoff’s law. White light emitted from core (photosphere) of the sun when passes through its atmosphere (chromospheres) radiations of those wavelengths will be absorbed by the gases present there, which they usually emit (in emission spectrum) resulting in dark lines in the spectrum of sun.
Stefan’s law radiant energy emitted per unit area per second (or emissive power or intensity) of a black body is directly proportional to the fourth power of temperature.
ER ∝ T4 or ER = σT4
If the body is not perfectly black then
ER = eσT4 where σ = 5.67 x10-8 Wm-2 K-4
Energy radiated per second or radiant power
PR = eAσT4 or ∫Rλdλ αT4
Cooling by radiation if a body is at temperature T in an environment of temperature T0 (< T), then body loses energy by emitting radiations at a rate
P1 = eAσT4
And it receives energy by absorbing radiation at a rate
P2 = eAσT40
So the net rate of loss is p = p1 – p2
Or p = eA σ(T4 – T40)
When a body cools by radiation, that rate of cooling depends upon the following factors:
Nature of the radiating surface, that is, emissivity
Breather the emissivity faster will be the cooling
Area of the radiating surface more the surface area of the radiating surface, faster will be the cooling.
Mass of the radiating surface greater the mass of the radiating body slower will be the cooling.
Specific heat of the radiating body more the specific heat slower will be the cooling.
Temperature of radiating body higher the temperature faster will be the cooling.
Temperature of the surrounding lesser the temperature of the surrounding faster is the cooling
Newton’s law of cooling the rate of cooling is proportional to the temperature difference between body and the surroundings provided the temperature difference is not very large from the surroundings. That is,
Dθ/dt = - K (θ – θ0)
Or ∫_θ2^θ2¦?(dθ? / θ )– θ0 = ∫t0- Kdt
Loge (θ2 – θ2) / (θ1 – θ0) = -Kdt
Show the temperature vs time curve.
If time intervals are equal and successive then
(θ2 – θ0) /( θ1 – θ0) = (θ3 – θ0) /( θ1 – θ0)
Assuming temperature at t =0 is θ1, after first interval it is θ2 and after second equal interval of time it isθ3.
Solar constant the sun is a perfectly black body as it emits all possible radiations (e = 1) solar consent s is defined as the intensity of solar radiation at the surface of the earth. That is,
S = pR / 4πr2 = 4πr2 σT4 / 4πr2
R = 1.5 x 1011 km
(Distance between the sun and the earth R = 7 x 108 km (radius of the sun)
S = 2cal / cm2 min = 138 W/m2
Wien’s displacement law when a body is heated it emits all possible radiations. However, intensity of different wavelengths is different. According to wine’s law the product of wavelength (corresponding to maximum intensity of radiations) and temperature of the body in Kelvin id constant
λmT = b (constant)
b = 2.89 x 10-3 m – K
Planck’s law Planck assumed that electromagnetic radiations are not emitted or absorbed continuously but in energy associated with each photon is
E = hv
Where h is Planck’s constant (= 6.625 x 10-34 j – s). On the basis of quantum theory Planck showed that
ER(λ) = 2πhc2 / λ5 /1 [ehc / λKT – 1]
ExpertsMind.com - Heat Transfer Laws Assignment Help, Heat Transfer Laws Homework Help, Heat Transfer Laws Assignment Tutors, Heat Transfer Laws Solutions, Heat Transfer Laws Answers, Thermodynamics Assignment Tutors, Physics Help