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Heat transferred by conduction into the sole of a firewalker's foot on hot coals. If the callus thickness is 3.00 mm, density is 300 kg/m³, area of contact is 25.0 cm², coal temperature is 700ºC, and contact time is 1.00 s, what is the heat transfer?

a) 125 J
b) 250 J
c) 375 J
d) 500 J

User Fejta
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1 Answer

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Final answer:

The heat transferred by conduction into the sole of a firewalker's foot is approximately 125 J.

Step-by-step explanation:

To calculate the heat transferred by conduction into the sole of the firewalker's foot, we can use the formula Q = k * A * (ΔT / d) * t, where Q is the heat transfer, k is the thermal conductivity, A is the area of contact, ΔT is the temperature difference, d is the thickness of the callus, and t is the time in contact. First, convert the thickness of the callus from millimeters to meters: d = 3.00 mm / 1000 = 0.003 m. Next, use the formula to calculate the heat transfer:

Q = (k * A * ΔT * t) / d

Plugging in the given values:

Q = (k * 0.025 m² * (700 °C - 0 °C) * 1 s) / 0.003 m

Using the average thermal conductivity of wood, which is approximately 0.1 W/(m·K), we can calculate the heat transfer:

Q = (0.1 W/(m·K) * 0.025 m² * (700 °C - 0 °C) * 1 s) / 0.003 m

Simplifying the equation gives:

Q ≈ 125 J

Therefore, the correct answer is option a) 125 J.

User Masade
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