A thermos flask with evacuated space to reduce the heat losses havingsurfaces facing each other of emissivity 0.02. If contents of the flaskare at 380 K & the ambient temperature is 298 K. Compute the heatloss from the flask. If same effect is to be achieved by using insulatingmaterial of conductivity 0.04 w/m K. What would be the thicknessrequired
Question
A thermos flask with evacuated space to reduce the heat losses havingsurfaces facing each other of emissivity 0.02. If contents of the flaskare at 380 K & the ambient temperature is 298 K. Compute the heatloss from the flask. If same effect is to be achieved by using insulatingmaterial of conductivity 0.04 w/m K. What would be the thicknessrequired
Solution
To solve this problem, we need to use the Stefan-Boltzmann Law and the formula for heat conduction.
- Stefan-Boltzmann Law: The formula is Q = εσA(T1^4 - T2^4), where Q is the heat transfer, ε is the emissivity, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area, and T1 and T2 are the temperatures of the two surfaces.
Given: ε = 0.02, T1 = 380K, T2 = 298K. The surface area (A) is not given, so we'll assume it to be 1 m^2 for simplicity.
Substituting these values into the formula, we get:
Q = 0.02 * 5.67 * 10^-8 * 1 * (380^4 - 298^4) = 0.02 * 5.67 * 10^-8 * (20857472000 - 7890481000) = 0.02 * 5.67 * 10^-8 * 12966981000 = 14.7 W
So, the heat loss from the flask is 14.7 W.
- Heat Conduction: The formula is Q = kA(T1 - T2)/d, where Q is the heat transfer, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures of the two surfaces, and d is the thickness of the material.
Given: Q = 14.7 W, k = 0.04 W/mK, A = 1 m^2, T1 = 380K, T2 = 298K. We need to find d.
Rearranging the formula to solve for d, we get:
d = kA(T1 - T2)/Q = 0.04 * 1 * (380 - 298)/14.7 = 0.04 * 82/14.7 = 0.224 m or 22.4 cm
So, the thickness of the insulating material required to achieve the same effect would be 22.4 cm.
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