Two iron spheres with the same radius look identical from the outside, but one has a hollow cavity on the inside (see below). They are both heated to a temperature of 350° C and suspended by a thin cord from the ceiling of a vacuum chamber. The only way they can lose heat is by radiation. (a) Which sphere, if either, will radiate more watts of power? Explain your answer (b) Which sphere, if either, will cool down the fastest? Explain your answer.

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**Title: Heat Radiation from Spheres in a Vacuum Chamber**

**Introduction:**

Two iron spheres with the same radius are visually identical from the outside. However, one sphere contains a hollow cavity inside, as demonstrated in the diagrams below. Both spheres are heated to a temperature of 350°C and suspended by thin cords from the ceiling of a vacuum chamber. The only means by which they can lose heat is through radiation.

**Objective:**

This exercise seeks to explore and explain the differences in heat radiation between the two spheres: 

1. **Which sphere, if either, radiates more power?** 
2. **Which sphere, if either, cools down the fastest?** 

Please provide explanations for each answer.

**Diagram Explanation:**

- The upper section of the image shows two iron spheres labeled "a" and "b" hanging from cords.
- The lower section contains an "interior view" for each sphere:
  - Sphere "a" shows a cross-section with a hollow cavity.
  - Sphere "b" shows a solid cross-section.

**Discussion Points:**

- **Heat Radiation:**
  - The rate of radiation depends on the surface area and temperature but not on the volume. Since both spheres have the same outer surface area and are at the same temperature, they will radiate the same amount of power. 

- **Cooling Rate:**
  - The hollow sphere (sphere "a") has less mass than the solid sphere (sphere "b") due to the absence of material in the cavity. Thus, sphere "a" has less thermal energy to lose and will cool down faster despite radiating the same amount of power per unit time. 

Understanding these principles illustrates fundamental concepts of thermodynamics and radiative heat transfer.
Transcribed Image Text:**Title: Heat Radiation from Spheres in a Vacuum Chamber** **Introduction:** Two iron spheres with the same radius are visually identical from the outside. However, one sphere contains a hollow cavity inside, as demonstrated in the diagrams below. Both spheres are heated to a temperature of 350°C and suspended by thin cords from the ceiling of a vacuum chamber. The only means by which they can lose heat is through radiation. **Objective:** This exercise seeks to explore and explain the differences in heat radiation between the two spheres: 1. **Which sphere, if either, radiates more power?** 2. **Which sphere, if either, cools down the fastest?** Please provide explanations for each answer. **Diagram Explanation:** - The upper section of the image shows two iron spheres labeled "a" and "b" hanging from cords. - The lower section contains an "interior view" for each sphere: - Sphere "a" shows a cross-section with a hollow cavity. - Sphere "b" shows a solid cross-section. **Discussion Points:** - **Heat Radiation:** - The rate of radiation depends on the surface area and temperature but not on the volume. Since both spheres have the same outer surface area and are at the same temperature, they will radiate the same amount of power. - **Cooling Rate:** - The hollow sphere (sphere "a") has less mass than the solid sphere (sphere "b") due to the absence of material in the cavity. Thus, sphere "a" has less thermal energy to lose and will cool down faster despite radiating the same amount of power per unit time. Understanding these principles illustrates fundamental concepts of thermodynamics and radiative heat transfer.
### Formulas:

#### Specific Heat:
- \( c \equiv \frac{Q}{m \Delta T} \) (J/kg · °C)
- \( Q = mc(T_f - T_i) \)

#### Mixtures:
- \( Q_{\text{cold}} = -Q_{\text{hot}} \)

#### Latent Heat:
- Phase change: \( Q = \pm mL \)

#### Thermal Conductivity:
- \( \frac{Q}{t} = \text{Power} = P = kA\frac{(T_h - T_c)}{L} \)

#### Stefan's Law of Radiation:
- \( P = \sigma A e T^4 \)
Transcribed Image Text:### Formulas: #### Specific Heat: - \( c \equiv \frac{Q}{m \Delta T} \) (J/kg · °C) - \( Q = mc(T_f - T_i) \) #### Mixtures: - \( Q_{\text{cold}} = -Q_{\text{hot}} \) #### Latent Heat: - Phase change: \( Q = \pm mL \) #### Thermal Conductivity: - \( \frac{Q}{t} = \text{Power} = P = kA\frac{(T_h - T_c)}{L} \) #### Stefan's Law of Radiation: - \( P = \sigma A e T^4 \)
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