Let's solve **question 4** step by step: --- ## Step-by-Step Solution **Given:** - Temperature at \( x = X \): \( T_X = 373 \) K - Ambient temperature: \( T_\infty = 25^\circ \mathrm{C} = 298 \) K - Base temperature (contact point): \( T_b = 250^\circ \mathrm{C} = 523 \) K - \( m = .1 \) - Find: \( X \) ### 1. **Equation for an Infinitely Long Fin** The temperature distribution along the fin is: \[ \frac{T(x) - T_\infty}{T_b - T_\infty} = e^{-mx} \] Where: - \( T(x) \): Temperature at distance \( x \) - \( T_\infty \): Ambient temperature - \( T_b \): Base temperature - \( m \): Parameter (given) - \( x \): Distance from base --- ### 2. **Plug in the Values** At \( x = X \): \[ \frac{T_X - T_\infty}{T_b - T_\infty} = e^{-mX} \] Plug in the numbers: - \( T_X = 373 \) K - \( T_\infty = 298 \) K - \( T_b = 523 \) K - \( m = .1 \) \[ \frac{373 - 298}{523 - 298} = e^{-.1X} \] \[ \frac{75}{225} = e^{-.1X} \] \[ \frac{1}{3} = e^{-.1X} \] --- ### 3. **Solve for \( X \)** Take natural logarithm on both sides: \[ \ln\left(\frac{1}{3}\right) = -.1X \] \[ - \ln(3) = -.1X \] \[ \ln(3) = .1X \] \[ X = \frac{\ln(3)}{.1} \] \[ X \approx \frac{1.0986}{.1} \] \[ X \approx 10.99 \] --- ## **Final Answer** \[ \boxed{X \approx 10.99\ \text{m}} \] --- ### **Explanation in English** - We used the temperature profile equation for an infinitely long fin. - Plugged in the known temperatures and solved for the unknown \( X \). - The result shows the distance from the fin base where the temperature is 373 K. --- If you need the answer for question 5 or further explanation, please let me know!