The question is: **"What is the percentage reduction in heat transfer if 7 shields of the same emissivities as the bodies are placed in between them?"** Let's solve it step by step: ### Heat Transfer with Radiation Shields For two surfaces with \( n \) shields placed between them, the heat transfer reduction can be calculated as follows: The heat transfer without shields (just two plates) is: \[ Q_0 = \frac{\sigma A (T_1^4 - T_2^4)}{\frac{2}{\epsilon} - 1} \] With \( n \) shields of the same emissivity \(\epsilon\), the formula is: \[ Q_n = \frac{\sigma A (T_1^4 - T_2^4)}{\left[ (n+1) \left(\frac{2}{\epsilon} - 1 \right) \right]} \] But more practically, the reduction ratio is: \[ \text{Reduction Ratio} = \frac{1}{n+1} \] So, the percentage reduction is: \[ \text{Percentage Reduction} = \left(1 - \frac{1}{n+1}\right) \times 100\% \] For **7 shields**: \[ n = 7 \] \[ \text{Percentage Reduction} = \left(1 - \frac{1}{7+1}\right) \times 100\% = \left(1 - \frac{1}{8}\right) \times 100\% = \left(\frac{7}{8}\right) \times 100\% = 87.5\% \] ### **Final Answer**: **87.5%** So, the correct answer is: - **87.5%**