A cube-shaped object with a side length of 17 cm has an emissivity of 0.5. The object is heated up to 322 0C and then placed in a room with a surrounding temperature of 36 0CUsing the Stefan-Boltzmann law, calculate the magnitude of the net power emitted by the object through radiation.Give your answer rounded to 4 decimal places without units.

Question

BlackTom AI · Accepted Answer

Stefan–Boltzmann law: P_net = ε σ A (T^4 − T_surr^4).

1) Relevant knowledge points and concepts
- Stefan–Boltzmann law for net radiative power between an object and surroundings: P_net = ε σ A (T^4 − T_env^4).
- Temperatures must be in kelvin. σ (Stefan–Boltzmann constant) = 5.670374419 × 10^−8 W·m^−2·K^−4.
- Surface area A of a cube = 6 a^2 (six faces).

2) Complete reasoning process step by step
- Convert all given quantities to SI:
  a = 17 cm = 0.17 m.
  T = 322 °C = 322 + 273.15 = 595.15 K.
  T_env = 36 °C = 36 + 273.15 = 309.15 K.
  ε = 0.5, σ = 5.670374419×10^−8 W·m^−2·K^−4.
- Compute area:
  A = 6 a^2 = 6 × (0.17 m)^2 = 6 × 0.0289 m^2 = 0.1734 m^2.
- Compute T^2 and T_env^2 (to help find T^4 − T_env^4):
  T^2 = 595.15^2 = 354,203.5225 K^2.
  T_env^2 = 309.15^2 = 95,573.7225 K^2.
- Use identity T^4 − T_env^4 = (T^2 − T_env^2)(T^2 + T_env^2):
  T^2 − T_env^2 = 354,203.5225 − 95,573.7225 = 258,629.8000.
  T^2 + T_env^2 = 354,203.5225 + 95,573.7225 = 449,777.2450.
  Thus T^4 − T_env^4 = 258,629.8 × 449,777.245 ≈ 1.16325798618901 × 10^11 K^4.
- Multiply through:
  σ A = (5.670374419×10^−8) × 0.1734 ≈ 9.8324292423 × 10^−9.
  σ A (T^4 − T_env^4) ≈ 9.8324292423×10^−9 × 1.16325798618901×10^11 ≈ 1,143.765186.
  Multiply by emissivity ε = 0.5: P_net = 0.5 × 1,143.765186 ≈ 571.882593.

3) Intermediate calculation summary (key numbers)
- A = 0.1734 m^2
- T^4 − T_env^4 ≈ 1.16325798618901×10^11 K^4
- σ A (T^4 − T_env^4) ≈ 1,143.765186
- P_net ≈ 0.5 × 1,143.765186 = 571.882593

4) Why the correct answer is correct
The value follows directly from correct application of the Stefan–Boltzmann law with proper conversion to kelvin, full cube surface area (all six faces radiate), inclusion of emissivity, and subtracting the environment’s radiative power. All arithmetic steps above lead to the computed net power.

5) Why common incorrect alternatives would be wrong
- Using Celsius instead of kelvin for T^4 would produce a large error.
- Using area of a single face (a^2) instead of 6a^2 would undercount by a factor of 6.
- Omitting the (T_env^4) subtraction would overestimate net emission.
- Forgetting emissivity (ε=1 vs 0.5) would double the result.

6) Conclusion confirming the final answer
Final net radiated power (rounded to four decimal places): 571.8826, matching the provided answer.

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