Determine the ratio of the heat flow into a six-pack of...

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - $\mathrm{L}$ plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of $6.00 \mathrm{~cm}$, a height of $12.0 \mathrm{~cm}$, and a thickness of $0.100 \mathrm{~cm}$. Use $205 \mathrm{~W} /(\mathrm{m} \mathrm{K})$ as the thermal conductivity of aluminum. Assume that the 2.00 - $\mathrm{L}$ bottle of soda has a diameter of $10.0 \mathrm{~cm}$, a height of $25.0 \mathrm{~cm}$, and a thickness of $0.100 \mathrm{~cm} .$ Use $0.100 \mathrm{~W} /(\mathrm{mK})$ as the thermal conductivity of plastic.

Key Concepts

Thermal Conductivity

Thermal conductivity is a material property that defines how well heat is transmitted through a material. It indicates the rate at which thermal energy is conducted through a unit area per unit temperature gradient, making it crucial for comparing the heat transfer abilities of different materials.

Fourier's Law of Heat Conduction

Fourier's Law describes the fundamental relationship governing heat conduction, stating that the heat transfer rate through a material is proportional to the thermal conductivity, the area through which the heat flows, and the temperature difference, while being inversely proportional to the thickness of the material. This law provides the foundation for calculating heat flow in various systems.

Geometric Considerations in Conduction

The geometry of an object, including dimensions such as area, thickness, and shape, plays a significant role in determining the rate of heat conduction. Larger surface areas promote increased heat transfer, while thicker materials tend to resist heat flow, making geometric analysis essential in problems involving conduction.

Heat Transfer Rate Comparison

Comparing heat transfer rates between different objects involves analyzing how variations in material properties and geometric parameters affect the overall conduction. By considering factors like thermal conductivity, surface area, and thickness, one can evaluate the relative effectiveness of different designs or materials in transferring heat.

Transcript

00:01 So the surface area of a cylinder can be given by 2 pi times d over 2, diameter over 2 squared plus pi times the diameter times h.

00:15 Or we can say that this would then be equalling to pi over 2 d squared plus pi d d h.

00:42 Now the heat transfer through the six soda cans, we can say the heat transfer to the soda cans would be equaling six times the thermal conductivity of the soda can times the surface area of the soda can multiplied by the change in temperature of the soda can divided by the length and so this would be equal to 6 k sub s delta t sub s divided by l sub s multiplied by ls of s multiplied by pi d squared over 2 plus pi d h and for the plastic bottle we can say p sub p this would be equaling the thermal conductivity of the plastic times the surface area of the plastic bottle times the change in temperature of the plastic bottle divided by the length of the plastic bottle and this would then be equal to k -sup p delta t sub p by l sub p and this would be multiplied by then pi d to p squared over two plus pi d sub p h sub p we know that the temperature difference you say delta p delta t sub p is equaling delta t sub s for the soda cans and so to find the ratio we can we can that then p sub s divided by p .p.

02:28 Would be equaling then after some algebraic manipulation, 6 k sub s, and say d sub s squared over 2 plus d sub s h sub s.

02:53 This would be divided by k sub p, multiplied by d sub p squared over 2 plus d sub p, h.

03:06 And at this point we can actually solve.

03:11 So this would be equaling to then 6, multiplied by 205 watts per meter per kelvin, multiplied then by 0 .06 meters, 6 centimeters, quantity squared divided by 2 plus point, 0 .06 meters multiplied by 0 .12 meters or 12 centimeters.

03:42 This would be divided then by 0 .10 watts per meter per kelvin.

03:53 Multiplied then by 0 .01 meters...

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