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A thin copper rod, 4 m in length, is heated at its midpoint and the ends are held at a constant temperature of [tex]$0^{\circ}$[/tex]. When the temperature reaches equilibrium, the temperature profile is given by [tex]$T(x) = 40 x (4 - x)$[/tex], where [tex]$0 \leq x \leq 4 \, \text{m}$[/tex] is the position along the rod. The heat flux at a point on the rod equals [tex][tex]$-k T^{\prime}(x)$[/tex][/tex], where [tex]$k$[/tex] is a constant. If the heat flux is positive at a point, heat moves in the positive [tex]$x$[/tex]-direction at that point, and if the heat flux is negative, heat moves in the negative [tex][tex]$x$[/tex][/tex]-direction.

a. With [tex]k=2[/tex], what is the heat flux at [tex]x=1[/tex]? At [tex]x=3[/tex]?
b. For what values of [tex]x[/tex] is the heat flux negative? Positive?
c. Explain the statement that heat flows out of the rod at its ends.

a. With [tex]k=2[/tex], the heat flux at [tex]x=1[/tex] is [tex]\square[/tex].
(Simplify your answer.)



Answer :

GinnyAnswer
Sure, let's break down the problem step-by-step:

### Part (a): Heat Flux at Specific Points

For a given temperature profile [tex]\( T(x) = 40x(4 - x) \)[/tex] and heat flux defined as [tex]\( q(x) = -k T'(x) \)[/tex]:

1. Finding [tex]\( T'(x) \)[/tex]:
[tex]\[ T(x) = 40x(4 - x) \][/tex]
Let's differentiate [tex]\( T(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ T'(x) = 40 \left( \frac{d}{dx} [x(4 - x)] \right) = 40 (4 - 2x) \][/tex]

2. Heat Flux Definition:
The heat flux at a point [tex]\( x \)[/tex] is given by [tex]\( q(x) = -k T'(x) \)[/tex].

3. Given [tex]\( k = 2 \)[/tex], calculate heat flux at specific points:
- At [tex]\( x = 1 \)[/tex]:
[tex]\[ T'(1) = 40(4 - 2 \cdot 1) = 40(4 - 2) = 40 \cdot 2 = 80 \][/tex]
[tex]\[ q(1) = -k T'(1) = -2 \cdot 80 = -160 \][/tex]
So, the heat flux at [tex]\( x = 1 \)[/tex] is [tex]\( -160 \)[/tex].

- At [tex]\( x = 3 \)[/tex]:
[tex]\[ T'(3) = 40(4 - 2 \cdot 3) = 40(4 - 6) = 40 \cdot (-2) = -80 \][/tex]
[tex]\[ q(3) = -k T'(3) = -2 \cdot (-80) = 160 \][/tex]
So, the heat flux at [tex]\( x = 3 \)[/tex] is [tex]\( 160 \)[/tex].

### Part (b): Values of [tex]\( x \)[/tex] for Negative and Positive Heat Flux

1. Heat Flux Expression:
[tex]\[ q(x) = -k T'(x) = -2 \cdot 40 (4 - 2x) = -80 (4 - 2x) \][/tex]

2. Analyzing the Heat Flux:
- Set [tex]\( 4 - 2x = 0 \)[/tex] to find the critical point:
[tex]\[ 4 - 2x = 0 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2 \][/tex]

3. Intervals for Heat Flux:
- For [tex]\( x < 2 \)[/tex]:
[tex]\[ 4 - 2x > 0 \Rightarrow q(x) = -80 (positive) \Rightarrow \text{heat flux is negative}. \][/tex]
- For [tex]\( x > 2 \)[/tex]:
[tex]\[ 4 - 2x < 0 \Rightarrow q(x) = -80 (negative) \Rightarrow \text{heat flux is positive}. \][/tex]

### Results for Heat Flux:
- Negative flux interval: [tex]\( (2, 4) \)[/tex]
- Positive flux interval: [tex]\( (0, 2) \)[/tex]

### Part (c): Heat Flow at Ends

- At [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex], the temperature [tex]\( T(x) = 0 \)[/tex], implying no temperature difference to drive the heat flux.
- However, since the heat flux at [tex]\( x = 0^+ \)[/tex] (immediately after [tex]\( 0 \)[/tex]) is positive and at [tex]\( x = 4^- \)[/tex] (immediately before [tex]\( 4 \)[/tex]) is negative, this indicates heat moves towards the ends of the rod from within. Therefore, as heat reaches the ends, it effectively flows out of the rod from both ends.

### Summary of Answers

a. Heat flux at [tex]\( x = 1 \)[/tex] is [tex]\(-160\)[/tex] and at [tex]\( x = 3 \)[/tex] is [tex]\(160\)[/tex].

b. Heat flux is negative for [tex]\( x \)[/tex] in the interval [tex]\( (2, 4) \)[/tex] and positive for [tex]\( x \)[/tex] in the interval [tex]\( (0, 2) \)[/tex].

c. This means that heat flows towards both ends of the rod and then leaves the rod at those points, consistent with heat moving to areas of lower temperature and eventually exiting the system.

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