Question

Entropy and the Sun The surface of the Sun has a temperature of 5500∘C and the temperature of deep space is 3.0K.Find the entropy increase produced by the Sun in one day, given that it radiates heat at the rate of 3.80 ×1026W

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Part B

How much work could have been done if this heat had been used to run an ideal heat engine?

Answer 1

Part - A -

Given that sun temperature = 5500 deg C

Convert this in kelvin,

T1 = 5500+273 K >> source >>

now, losses heat in 1 day = Q1
Q1 = 3.80x10^26 (W)*(24*3600 sec) Joule
Q1 = 3283.2*10^28 J

Again, deep space >> sink temp (T2 = 3 K) >>
gains heat in 1 day = Q1
change in entropy dS = dQ/T >> defined as heat gained/lost per unit of temp (deg K) at constant temp.
ds = + ve when heat gained (disorder increases)
ds = - ve when heat lost (disorder decreases)
net change in entropy = dS = - Q1/T1 + Q1/T2
dS = Q1 [T1 - T2] / T1 T2] = Q1[5770/5773*3]
dS = 3283.2*10^28 *0.333
dS = + 1093.34*10^28 J/deg K
dS = 10.93*10^30 J/K

therefore, entropy increase produced by the sun in one day = 10.93 x 10^30 J/K

Part - B -

Now the work done for ideal pump -
Efficiency of pump = (T1-T2)/T1 = Work done/Q1

T1 = 5500 + 273 = 5773 K

T2 = 3K

put the values-

W = Q1 *[5770/5773] = (3283.2 x 10^28)*[5770 / 5773] = 3281.5 x 10^28 = 32.81 x 10^30 Joule.

So, the work done if this heat had benn used to run an ideal heat engine = 32.81 x 10^30 Joule.