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A person eats a dessert that contains 220 calories. (this "calorie unit, with a capital c, is the one used by nutritionists; 1 calorie = 4186 j). the skin temperature of the person is 37 degrees celsius and that of her environment is 24 degrees celsius. the emissivity of her skin is 0.69 and its surface area is 1.0 m^2. how many hours would it take her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

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Final answer:

To find the time it would take for the person to emit a net radiant energy equal to the energy in the dessert, we need to calculate the rate of heat transfer by radiation using the formula Q = εσA(T₁⁴ - T₂⁴). We can then divide the energy in the dessert by this rate of heat transfer to find the time.

Step-by-step explanation:

To calculate the time it would take for the person to emit a net radiant energy equal to the energy in the dessert, we need to calculate the rate of heat transfer by radiation. Using the equation for the rate of radiative heat transfer, we can determine the rate of heat transfer per unit time. We then divide the energy contained in the dessert by this rate of heat transfer to find the time.

The formula for the rate of heat transfer by radiation is given by: Q = εσA(T₁⁴ - T₂⁴), where Q is the rate of heat transfer, ε is the emissivity of the skin, σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²K⁴), A is the surface area of the skin, and T₁ and T₂ are the temperatures of the skin and the environment, respectively. In this case, T₁ = 37°C = 310 K and T₂ = 24°C = 297 K.

Plugging in the values, we get: Q = (0.69)(5.67 × 10⁻⁸ W/m²K⁴)(1.0 m²)(310⁴ - 297⁴).

Next, we rearrange the equation to solve for Q and find: Q = (0.69)(5.67 × 10⁻⁸ W/m²K⁴)(1.0 m²)[(310⁴) - (297⁴)].

Now, we calculate the energy in the dessert. Since 1 calorie is equal to 4186 J, the energy in the dessert is 220 calories × 4186 J/calorie = 920,920 J.

To find the time, we divide the energy in the dessert by the rate of heat transfer: time = 920,920 J / Q.

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