Let's solve the problem step-by-step: **Given:** - Thickness of slab (\(L\)) = **10 mm** = **0.01 m** - Thermal conductivity (\(k\)) = **15 W/m·K** - Heat transfer coefficient at left (\(h_1\)) = **22 W/m²·K** - Heat transfer coefficient at right (\(h_2\)) = **18 W/m²·K** The **overall heat transfer coefficient** (\(U\)) is given by: \[ \frac{1}{U} = \frac{1}{h_1} + \frac{L}{k} + \frac{1}{h_2} \] Substitute the values: \[ \frac{1}{U} = \frac{1}{22} + \frac{0.01}{15} + \frac{1}{18} \] Calculate each term: - \(\frac{1}{22} \approx 0.0455\) - \(\frac{0.01}{15} \approx 0.000667\) - \(\frac{1}{18} \approx 0.0556\) Sum them: \[ \frac{1}{U} = 0.0455 + 0.000667 + 0.0556 = 0.101767 \] Now, invert to find \(U\): \[ U = \frac{1}{0.101767} \approx 9.82 \text{ W/m}^2\text{K} \] **Closest option:** **9.84** is the closest to our calculation. ### **Final Answer:** **Option 2: 9.84**