Two thermal reservoirs are connected by a solid copper bar....

Two thermal reservoirs are connected by a solid copper bar. The bar is $2.00 \mathrm{~m}$ long, and the temperatures of the reservoirs are $80.0^{\circ} \mathrm{C}$ and $20.0^{\circ} \mathrm{C}$ a) Suppose the bar has a constant rectangular cross section, $10.0 \mathrm{~cm}$ on a side. What is the rate of heat flow through the bar? b) Suppose the bar has a rectangular cross section that gradually widens from the colder reservoir to the warmer reservoir. The area $A$ is determined by $A=\left(0.0100 \mathrm{~m}^{2}\right)[1.0+x /(2.0 \mathrm{~m})],$ where $x$ is the distance along the bar from the colder reservoir to the warmer one. Find the heat flow and the rate of change of temperature with distance at the colder end, at the warmer end, and at the middle of the bar.

Two thermal reservoirs are connected by a solid copper bar. The bar is $2.00 \mathrm{~m}$ long, and the temperatures of the reservoirs are $80.0^{\circ} \mathrm{C}$ and $20.0^{\circ} \mathrm{C}$
a) Suppose the bar has a constant rectangular cross section, $10.0 \mathrm{~cm}$ on a side. What is the rate of heat flow through the bar?
b) Suppose the bar has a rectangular cross section that gradually widens from the colder reservoir to the warmer reservoir. The area $A$ is determined by $A=\left(0.0100 \mathrm{~m}^{2}\right)[1.0+x /(2.0 \mathrm{~m})],$ where $x$ is the distance
along the bar from the colder reservoir to the warmer one. Find the heat flow and the rate of change of temperature with distance at the colder end, at the warmer end, and at the middle of the bar.

Key Concepts

Variable Cross-Sectional Area in Heat Conduction

When the cross-sectional area of a conducting material changes along its length, the analysis of heat transfer requires integration to account for the varying contributions of different sections. This concept generalizes the simple application of Fourier’s law to more complex geometries, enabling the calculation of heat flow and temperature gradients at different locations along the material.

Temperature Gradient

The temperature gradient is the rate at which temperature changes with respect to distance. In conduction problems, it is used to determine how efficiently heat is transferred through a material and is a key part of the relationship outlined in Fourier’s law.

Thermal Conductivity

Thermal conductivity is a material property that measures a material's ability to conduct heat. In problems involving heat transfer, knowing the thermal conductivity allows one to relate the temperature gradient across a material with the rate of heat transfer, making it crucial for understanding how quickly or slowly heat moves through substances.

Fourier’s Law of Heat Conduction

Fourier’s law is the fundamental principle governing the conduction of heat. It states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the cross-sectional area perpendicular to the heat flow. This law applies both to systems with uniform cross sections and those with geometries that vary along the direction of heat flow.

Steady-State Heat Conduction

Steady-state heat conduction refers to a scenario where the temperature at any given point in a material does not change over time, meaning the heat flowing into any segment of the material is equal to the heat flowing out. This assumption simplifies the analysis, as it allows the use of constant heat flux when applying Fourier’s law.

Transcript

00:01 So in this question we have length of the bar l equals to 2 meter and area of the copper bar a equals to 10 cm, mudblared by 10 cm that is 100 cm square that is 0 .01 meter square okay and the temperature of the hot end th equals to 100 degrees centigrade and temperature of the cold end tc equals to 20 degrees centigrade and temperature of the cold end tc equals to 20 degrees centigrade and also also we know that the thermal conductivity of the copper kcu copper equals to 386 watt per meter calvin.

00:41 Now the rate of heat flow through the copper bar is given by p equals to k copper made by a, t hot minus t cold divided by l.

00:54 This is our equation number one.

00:56 Okay.

00:58 And that rate of heat flow along the length of the bar can be obtained by the following equation.

01:06 This is the equation for constant values and we can also write k copper a multiplied by delta t by delta x when delta x varies with length.

01:21 Okay.

01:22 So from here we can write that so from here we can write that delta t by delta x that is equals to p by k copper multiplied by area a and the area is related to the length as a equals to 0 .01 meter square 1 plus x by 2 meter okay, so substituting this area in this equation we get p equals to so delta t by delta x equals to delta t by delta x equals to p divided by k copper but plowed by 0 .011 plus x by 2 meter okay so this is our equation number two now moving to the part a of the problem in which we have to determine the power.

02:24 So substituting the values in equation 1, we get rate of heat flow that is power equals to 386 multiplied by 0 .01 multiplied by 100 degrees centigrade minus 20 degrees centigrade divided by 2 meter.

02:41 So from here after solving we get p equals to 115 .8 watt that is nearly equals to 116 6 .0 watt.

02:52 So this is the answer for the part a of the problem.

02:57 Rate of heat flow.

02:58 Now moving to the part b in which we have to determine the rate of heat flow at the cold end that is x equals to 0.

03:07 So substituting value of x in the equation number 2 as we have determined above.

03:12 So delta t by delta x will be at x equals to 0 that will be equals to p by p is 115 .8 watt this value divided by k that is 386 bracket 0 .01 again bracket 1 plus x by 2 that is x equals to 0 so 0 by 2 okay so from here after solving we get delta t by delta x that is temperature gradient at cold and x equals to 0 comes out to be 30 .0 .30 .0 degrees centigrade per meter okay so now this is the answer and now at the middle of the bar that is as x equals to one meter we'll get delta t by delta x at x equals to 1 this will be equals to 115 .8 divided by 386 bracket 0 .01 again bracket 1 plus 1 by 2 so from here after solving we will get delta t by delta x at middle of the bar that is x equals to 1.

04:30 This will be equal to 20 .0 degree centigrade per meter...

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