The calculation of the heat transfer occurs through the following formula:

**Q = -kAdT/dx = – kA*(T2 – T1)/dx = – kA*(T2 – T1)/L**

In the above formula, * k* is the heat transfer constant of the material,

**and**

*T1***represent the initial and final temperatures respectively while**

*T2**is the surface area of the material and*

**A***represents the length. For the transfer of heat between*

**L***, the solution appears as follows:*

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**QQ**

= – kA*(T2 – T1)/L

= -2*(5*7.5) *(1000 – 600)/1

= -2*37.5*400/1

= -30,000

**QB** = – kA*(T2 – T1)/L

= -4*(5*7.5) *(1000 – 600)/5

= -150*400/5 = -12,000

**QC** = – kA*(T2 – T1)/L

= -8(5*2) *(1000 – 600)/10

= – 3,200

**Q**D = – kA*(T2 – T1)/L

= -15*(5*1.5) *(100 – 600)/10

= – 4,500

**QE** = – kA*(T2 – T1)/L

= – 20*(5*4) *(1000 – 400)/10

= -16,000

**QF** = – kA*(T2 – T1)/L

= -35*(5*7.5) *(1000 – 600)/5

= -105,000

The heat transfer through the materials was different. It varied from material to material due to various reasons. First, the heat transfer depends on the constant of heat conductivity which depends on each material. For example, metals have high constant of conductivity K than other materials such as wood or concrete. The heat transfer further varies from one metal to the other.

Another factor that contributed to the variation in the heat transfer among the materials is the surface area. When a material has a large surface area, it absorbs more heat leading to faster heat transfer than materials that have low surface areas. Based on the findings, the materials that had smaller surface areas exposed to the source of heat exhibited lower rates of transfer.

Finally, the variation depended on the length of the transfer. When the heat has to cover a long distance, it will take a longer time to travel over the material as opposed to the transfer along a shorter material. Thus, when the temperature difference is kept constant, the heat conductivity constant, area, and length are the three primary determinants of the rate of heat transfer in a material.

**From To T1 (0C) T2 (0C) LF(m) Dt**

0 10 1000 600 0.15 400

11 20 850 550 0.12 300

21 30 750 500 0.1 250

31 40 600 400 0.08 200

41 50 500 350 0.07 150

51 60 300 200 0.06 100

61 70 300 100 0.05 200

71 80 200 100 0.04 100

81 90 200 50 0.08 150

91 99 200 100 0.015 100

A plot of the temperature change against the length of the material produced a positive slope showing a the existence of a positive correlation between the two constants. The positive slope is an indication of the heat conductivity constant which is always positive. The formula for finding the heat transfer across a medium is – kA*(T2 – T1)/L where k is the constant of conductivity. Hence, the slope of 2272 represents the constant.

The linear regression resulted in an equation y = 2272.8x + 21.13. In the equation, 21.13 shows the room temperature from which the transfer commences. The slope of the graph is therefore 2272.8 which is the constant of conductivity of a material. It shows the rate at which the material will transfer heat given other factors remain constant. The rate is therefore positive showing as it defines the speed at which a material will transfer the heat energy from one point to another.

**Section B**

An example of a mass diffusion process is the food coloring that occurs in the industries. Most of these industries have automated processes that manufacture tons of foods in a day. The food manufacturing companies such as those producing drinks always rely on the natural diffusion process to mixture the food with the food coloring agent to obtain a uniform color of its food products. The firms would not receive satisfactory response from the clients if they could produce food products that have varied coloring. For example, the Coca-Cola drink is known to be dark red in color. If the company would sell some pale colored soda, there is no client that would consume it for fear of having an unsafe product. Therefore, the natural diffusion process helps the company to achieve its coloring goal.

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