The rate of heat transfer by radiation between a body and its surroundings can be calculated using the Stefan-Boltzmann Law. The formula is:

$Q = \varepsilon \sigma A (T_b^4 - T_s^4)$

Where:

• $Q$ is the rate of heat transfer (W),
• $\varepsilon$ is the emissivity of the surface (dimensionless),
• $\sigma$ is the Stefan-Boltzmann constant, $\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4$,
• $A$ is the surface area of the body (m²),
• $T_b$ is the temperature of the body in kelvin (K),
• $T_s$ is the temperature of the surroundings in kelvin (K).

Step-by-step process:

1. Convert temperatures from Celsius to Kelvin: $T_b = 175 + 273.15 = 448.15 \, \text{K}$ $T_s = 25 + 273.15 = 298.15 \, \text{K}$

2. Plug the known values into the Stefan-Boltzmann equation: $Q = 0.75 \times 5.67 \times 10^{-8} \times 0.65 \times ((448.15)^4 - (298.15)^4)$

Let's compute this.The rate of heat transfer by radiation between the body and its surroundings is approximately $896.5 \, \text{W}$.

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does emissivity affect the rate of heat transfer by radiation?
2. What happens to the heat transfer if the surface area of the body increases?
3. How would the heat transfer change if the temperature of the surroundings increased?
4. Why is it necessary to convert temperatures to Kelvin when using the Stefan-Boltzmann law?
5. Can the heat transfer by radiation ever be negative?

Tip: In radiation problems, always ensure temperatures are converted to Kelvin to avoid incorrect calculations.