A 25-ohm resistor at steady state draws a current of 10.8 amperes. its temperature is 310 k; the temperature of the surroundings is 300 k. what is the total rate of entropy generation ?

Respuesta :

The total rate of entropy generation is 0.268 J/K.

Entropy, the degree of a gadget's thermal electricity consistent with unit temperature that is unavailable for doing useful paintings. due to the fact work is acquired from ordered molecular motion, the quantity of entropy is likewise a measure of the molecular sickness, or randomness, of a machine.

Resistance of resistor R = 25 ohm

the total rate of entropy generation ?

Current flowing through resistor I = 10 A

Heat generation due to resistance , Q = I2R = 100*25 = 2500 J

Resistor temperature T = 310 K

Surrounding temperature Ts = 300 K

Entropy change for Resistor, heat lost due to the resistance to the surrounding then Q = -ve

(ΔS)syst = - Q/T = - 2500 J/310K = - 8.064 J/K

Entropy change for the system: here heat gain from the resistor, Q = +ve

(ΔS)surr = +Q/Ts = 2500J/300K = 8.333 J/K

Total Entropy generation :

(ΔS)tot = (ΔS)syst + (ΔS) Surr = -8.064 + 8.333 = 0.268 J/K

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