Some cooking pan whose bottom is made out of copper has a radius 0.1 m and thickness 0.003 m. The water in the pan boils at 1.7e-3 kg/s. How much higher is temperature of the outside surface of the pan's bottom compared to the inside surface?
The heat transfer rate is
Power = 1.7 g/s * 540 cal/g
= 918 cal/s
= 3841 J/s
Power = k*A*dT/dx,
where k is the thermal conductivity of copper (which you need to look up) and A is the pan's bottom area, 0.03142 m^2.
Solve for the temperature gradient, dT/dx = Power/(k*A).
The temperature difference between inside and outside surfaces of the copper is the copper thickness x, multiplied by dT/dx.
T2 - T1 = x*Power/(k*A)
To determine the temperature difference between the outside and inside surface of the pan's bottom, we can use the formula for heat transfer through conduction.
The heat transfer rate (Q) through conduction can be calculated using the following equation:
Q = (k * A * (T1 - T2)) / d
where:
- Q is the heat transferred per unit time (in watts, W)
- k is the thermal conductivity of the material (in watts per meter-kelvin, W/(m·K))
- A is the surface area of the pan's bottom (in square meters, m^2)
- T1 is the temperature of the outside surface (in kelvin, K)
- T2 is the temperature of the inside surface (in kelvin, K)
- d is the thickness of the pan's bottom (in meters, m)
In this case, we need to find the temperature difference (T1 - T2). Rearranging the equation, we get:
(T1 - T2) = (Q * d) / (k * A)
Now, let's calculate the values needed to plug into the equation:
- The thermal conductivity (k) of copper is approximately 400 W/(m·K).
- The surface area (A) of the pan's bottom can be calculated using the formula for the area of a circle: A = π * r^2, where r is the radius of the pan's bottom.
- Given radius (r) = 0.1 m, so A = π * (0.1)^2 = 0.0314 m^2.
Substituting these values into the equation, we have:
(T1 - T2) = (Q * d) / (k * A)
(T1 - T2) = (1.7e-3 kg/s * 0.003 m) / (400 W/(m·K) * 0.0314 m^2)
Now, we can calculate the temperature difference:
(T1 - T2) = (5.1e-6 kg⋅m/s) / (0.01256 W/(K·m))
(T1 - T2) ≈ 406.36 K
Therefore, the temperature of the outside surface of the pan's bottom is approximately 406.36 K higher than the inside surface.