Question

The Sun radiates energy at the rate of 3.80×10²⁶ W from its 5500º C surface into dark empty space (a negligible fraction radiates onto Earth and the other planets). The effective temperature of deep space is −270º C. (a) What is the increase in entropy in one day due to this heat transfer? (b) How much work is made unavailable?

a) 2.88×10¹⁴ J/K, 0 J
b) 3.20×10¹⁴ J/K, 0 J
c) 3.52×10¹⁴ J/K, 0 J
d) 3.84×10¹⁴ J/K, 0 J

Answer 1

The increase in entropy in one day due to the Sun's energy radiated into space is approximately 3.52×10¹⁴ J/K, and the work made unavailable by this process is 0 J. This calculation is done assuming the temperatures are converted to Kelvin and using the power output of the Sun.

Step-by-step explanation:

Since the Sun radiates energy into space with a high temperature, and the effective temperature of deep space is significantly lower, this results in an increase in entropy. The entropy change (ΔS) due to the heat transfer (Q) over a temperature (T) is given by the formula ΔS = Q/T, assuming a reversible process. We must first convert temperatures from Celsius to Kelvin: T₁ (Sun's surface) = 5500 °C + 273.15 = 5773.15 K and T₂ (deep space) = -270 °C + 273.15 = 3.15 K.

To calculate the increase in entropy over one day (86,400 seconds), we use the Sun's radiated power (P = 3.80×10²⁶ W) and the temperature of deep space. Hence, ΔS = P×time/T₂ = 3.80×10²⁶ W × 86,400 s / 3.15 K, which calculates to an increase in entropy of approximately 3.52×10¹⁴ J/K, corresponding to option (c).

For part (b), since deep space is not able to perform work (it's at 0 Kelvin technically, but for our calculation purposes, it's 3 K), the work made unavailable due to this heat transfer is effectively 0 J.