QuestionDecember 14, 2025

70. II A cubical box 20 cm on a side is constructed from 1.2-cm- thick concrete panels. A 100 W lightbulb is sealed inside the box. What is the air temperature inside the box when the light is on if the surrounding air temperature is 20^circ C

70. II A cubical box 20 cm on a side is constructed from 1.2-cm- thick concrete panels. A 100 W lightbulb is sealed inside the box. What is the air temperature inside the box when the light is on if the surrounding air temperature is 20^circ C
70. II A cubical box 20 cm on a side is constructed from 1.2-cm- thick concrete panels. A 100 W lightbulb is sealed inside the box. What is the air temperature inside the box when the light is on if the surrounding air temperature is 20^circ C

Solution
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Answer

25^\circ\text{C} Explanation 1. Calculate surface area for heat transfer The box has 6 faces, each 20\,\text{cm} \times 20\,\text{cm} = 400\,\text{cm}^2 = 0.04\,\text{m}^2. Total area A = 6 \times 0.04 = 0.24\,\text{m}^2. 2. Find thermal resistance of one panel Thickness L = 1.2\,\text{cm} = 0.012\,\text{m}; concrete conductivity k \approx 1.0\,\text{W}/(\text{m}\cdot\text{K}). Resistance per face: R = \frac{L}{kA} = \frac{0.012}{1.0 \times 0.24} = 0.05\,\text{K/W}. 3. Relate power to temperature difference All heat escapes through panels: P = \frac{\Delta T}{R} \implies \Delta T = PR. P = 100\,\text{W}, R = 0.05\,\text{K/W}. 4. Calculate inside temperature \Delta T = 100 \times 0.05 = 5\,\text{K}. Inside temp = 20^\circ\text{C} + 5^\circ\text{C} = 25^\circ\text{C}.

Explanation

1. Calculate surface area for heat transfer<br /> The box has 6 faces, each $20\,\text{cm} \times 20\,\text{cm} = 400\,\text{cm}^2 = 0.04\,\text{m}^2$. Total area $A = 6 \times 0.04 = 0.24\,\text{m}^2$.<br />2. Find thermal resistance of one panel<br /> Thickness $L = 1.2\,\text{cm} = 0.012\,\text{m}$; concrete conductivity $k \approx 1.0\,\text{W}/(\text{m}\cdot\text{K})$. Resistance per face: $R = \frac{L}{kA} = \frac{0.012}{1.0 \times 0.24} = 0.05\,\text{K/W}$.<br />3. Relate power to temperature difference<br /> All heat escapes through panels: $P = \frac{\Delta T}{R} \implies \Delta T = PR$. $P = 100\,\text{W}$, $R = 0.05\,\text{K/W}$.<br />4. Calculate inside temperature<br /> $\Delta T = 100 \times 0.05 = 5\,\text{K}$. Inside temp $= 20^\circ\text{C} + 5^\circ\text{C} = 25^\circ\text{C}$.
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