A rubber ball of radius 2.12 m and emissivity 0.940 is heated to 138°C. HINT (a) Convert the given temperature to Kelvin. (Enter your answer to at least three significant figures.) 411 K (b) What is the surface area (in m2) of the ball? 56.47 (c) If the ambient temperature is 24.0°C, what net power (in W) does the ball radiate? 6.34E6 X W

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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please help with part C. Thanks :) answer in W

**Problem:**

A rubber ball of radius 2.12 m and emissivity 0.940 is heated to 138°C.

**Instructions:**

- Consider the hint as a resource for solving the problem.
  
**Solutions:**

(a) Convert the given temperature to Kelvin. (Enter your answer to at least three significant figures.)

- Answer: 411 K ✓

(b) What is the surface area (in m²) of the ball?

- Answer: 56.47 m² ✓

(c) If the ambient temperature is 24.0°C, what net power (in W) does the ball radiate?

- Answer: 6.34E6 W ✗

**Explanation:**

- For part (a), the temperature conversion is correct.
- For part (b), the surface area calculation is correct using the formula \(4\pi \times \text{radius}^2\).
- For part (c), there is an error with the net power calculation. It requires recalculation using the Stefan-Boltzmann Law formula: \(P = e\sigma A (T^4 - T_{\text{ambient}}^4)\). Make sure to use consistent units and recalculate as needed.
Transcribed Image Text:**Problem:** A rubber ball of radius 2.12 m and emissivity 0.940 is heated to 138°C. **Instructions:** - Consider the hint as a resource for solving the problem. **Solutions:** (a) Convert the given temperature to Kelvin. (Enter your answer to at least three significant figures.) - Answer: 411 K ✓ (b) What is the surface area (in m²) of the ball? - Answer: 56.47 m² ✓ (c) If the ambient temperature is 24.0°C, what net power (in W) does the ball radiate? - Answer: 6.34E6 W ✗ **Explanation:** - For part (a), the temperature conversion is correct. - For part (b), the surface area calculation is correct using the formula \(4\pi \times \text{radius}^2\). - For part (c), there is an error with the net power calculation. It requires recalculation using the Stefan-Boltzmann Law formula: \(P = e\sigma A (T^4 - T_{\text{ambient}}^4)\). Make sure to use consistent units and recalculate as needed.
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