determine the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water. The heat of vaporization of water at 100°C is 2257 kJ/kg. Answers: 46.2 W, 56.1 W, 0.082 Vapor 2 kg/h 25°C Water 98°C 100°C ɛ = 0.95

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Problem Statement:**

Determine the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water. The heat of vaporization of water at 100°C is 2257 kJ/kg.

**Answers:**
- Heat lost from side surfaces: 46.2 W
- Heat lost by evaporation: 56.1 W
- Ratio: 0.082

**Diagram Explanation:**

The diagram (Figure P9–22) illustrates a pan with water being heated. Key features include:

- **Water:** The water inside the pan is at 100°C.
- **Vapor:** Water vapor is shown rising from the surface at a rate of 2 kg/h.
- **Temperature Label:** The handle is labeled at 25°C.
- **Pan Surface:** The side of the pan is at 98°C with an emissivity (ε) of 0.95.

This setup is used to compare the heat loss mechanisms via the side surfaces and the evaporation process.
Transcribed Image Text:**Problem Statement:** Determine the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water. The heat of vaporization of water at 100°C is 2257 kJ/kg. **Answers:** - Heat lost from side surfaces: 46.2 W - Heat lost by evaporation: 56.1 W - Ratio: 0.082 **Diagram Explanation:** The diagram (Figure P9–22) illustrates a pan with water being heated. Key features include: - **Water:** The water inside the pan is at 100°C. - **Vapor:** Water vapor is shown rising from the surface at a rate of 2 kg/h. - **Temperature Label:** The handle is labeled at 25°C. - **Pan Surface:** The side of the pan is at 98°C with an emissivity (ε) of 0.95. This setup is used to compare the heat loss mechanisms via the side surfaces and the evaporation process.
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