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PtichkaEL [24]
3 years ago
14

Consider a blackbody that radiates with an intensity i1 at a room temperature of 300k. At what intensity i2 will this blackbody

radiate when it is at a temperature of 400k?
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

3.16 i1

Explanation:

T1 = 300 k

I1 = i1

T2 = 400 k

i2 = ?

The intensity is directly proportional to the four powers of the absolute temperature.

I ∝ T^4

So, \frac{i_{2}}{i_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}

\frac{i_{2}}{i_{1}}=\frac{400^{4}}{300^{4}}

i2 = 3.16 i1

Thus, the intensity becomes 3.16 i1.

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Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an i
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Answer:

1) the entropy generated is Δs= 0.0363 kJ/kg K

2) the minimum theoretical work is w piston = 201.219 kJ/kg

Explanation:

1) From the second law of thermodynamics applied to an ideal gas

ΔS = Cp* ln ( T₂/T₁) - R ln (P₂/P₁)

and also

k= Cp/Cv , Cp-Cv=R → Cp*( 1-1/k) = R → Cp= R/(1-1/k)= k*R/(k-1)

ΔS =  k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

where R= ideal gas constant , k= adiabatic coefficient of air = 1.4

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ΔS =  k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

ΔS = 1.4/(1.4-1) *8.314 J/mol K * ln( 600K/300K) - 8.314 J/mol K *  ln (10 bar/ 1bar)

ΔS = 1.026 J/ mol K

per mass

Δs = ΔS / M

where M= molecular weight of air

Δs = 1.026 J/ mol K / 28.84 gr/mol = 0.0363 J/gr K = 0.0363 kJ/kg K

2) The minimum theoretical work input is carried out under a reversible process. from the second law of thermodynamics

ΔS =∫dQ/T =0 since Q=0→dQ=0

then

0 = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

T₂/T₁ = (P₂/P₁)^[(k-1)/k]

T₂ = T₁ * (P₂/P₁)^[(k-1)/k]

replacing values

T₂ = 300K * ( 10 bar/1 bar)^[0.4/1.4] = 579.2 K

then from the first law of thermodynamics

ΔU= Q - Wgas = Q + Wpiston ,

where ΔU= variation of internal energy , Wgas = work done by the gas to the piston , Wpiston  = work done by the piston to the gas

since Q=0

Wpiston = ΔU

for an ideal gas

ΔU= n*Cv*(T final - T initial)

and also

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then

ΔU= n*R/(k-1)*(T₂  - T₁)

W piston = ΔU = n*R/(k-1)*(T₂  - T₁)

the work per kilogram of air will be

w piston = W piston / m = n/m*R/(k-1)*(T₂  - T₁)  = (1/M*) R/(k-1)*(T₂  - T₁)  ,

replacing values

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