(a)
A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction (in J) into the sole of one foot of a firewalker given that the bottom of the foot is a 2.90 mm-thick callus with a conductivity at the low end of the range for wood and its density is 300 kg/m3. The area of contact is 25.0 cm2, the temperature of the coals is 700°C, and the time in contact is 1.06 s.
48.5
J
(b)
What temperature increase (in degrees C) is produced in the 7.25 cm3 of tissue affected?
Now wait a minute, I just did the pan with the water. Do this pretty much the same way.
To calculate the heat transferred by conduction, we can use the equation:
Q = k*A*(deltaT/deltaX)*t
Where:
Q = Heat transferred (in Joules)
k = Thermal conductivity of the material (in W/m∙K)
A = Area of contact (in m²)
deltaT = Temperature difference (in K)
deltaX = Thickness of the material (in m)
t = Time in contact (in seconds)
For part (a), we need to find the heat transferred into the sole of one foot. Given the thickness of the callus (deltaX) is 2.90 mm (0.0029 m), the conductivity (k) is at the low end of the range for wood (we can assume k = 0.15 W/m·K), the density of the callus (p) is 300 kg/m³, the area of contact (A) is 25.0 cm² (0.0025 m²), the temperature difference (deltaT) is the difference between the temperature of the coals and the initial temperature (let's assume 0°C), and the time in contact (t) is 1.06 s.
First, let's convert the thickness of the callus to meters:
deltaX = 0.0029 m
Now, let's calculate the volume of the callus:
Volume = Area * Thickness
Volume = 0.0025 m² * 0.0029 m
Volume = 7.25e-6 m³
To find the mass of the callus, we can use the density:
Mass = Density * Volume
Mass = 300 kg/m³ * 7.25e-6 m³
Mass = 0.002175 kg
Now, let's find the temperature difference:
deltaT = 700°C - 0°C
deltaT = 700 K
Finally, we can calculate the heat transferred:
Q = k * A * (deltaT/deltaX) * t
Q = 0.15 W/m·K * 0.0025 m² * (700 K / 0.0029 m) * 1.06 s
Q ≈ 48.5 J
Therefore, the heat transferred by conduction into the sole of one foot of the firewalker is approximately 48.5 J.