CH 13 - 1410.002 Question 5 of 10 View Policies Current...

CH 13 - 1410.002 Question 5 of 10 View Policies Current Attempt in Progress Two objects are maintained at constant temperatures, one hot and one cold. Two identical bars can be attached end to end, as in part a of the drawing, or one on top of the other, as in part b. When either of these arrangements is placed between the hot and the cold objects for the same amount of time, heat $Q$ flows from left to right. Find the ratio $Q_a/Q_b$. (a) (b) $Q_a/Q_b$ = Number Units No units m °F kg J °C W Attempts: 0 of 5 used Submit Answer

Transcript

00:01 For this problem, we have two type of arrangements for two identical bars.

00:10 They could be one top of the other as in part as this part e, or one side by the other as in part a.

00:24 Now, we know that regardless of the arrangement, the heat flows from left to right.

00:36 So, and the first question of this problem is which heat flows, what is the area, the cross -sectional area through which the heat floats is greater in part in the configuration a or in the configuration b.

00:59 So that's the first problem.

01:02 So taking into account the thing, what it says in the problem that flow, that he always flows to left to right, from that to right.

01:22 We need to know that we need to take into account these cross -sectional area.

01:32 And we can write this, we can easily see this, if we can write the other area of a should be, we are going to put numbers in these sides.

01:42 So we call this d and we call these l.

01:47 We call this l and we call this d prime.

01:53 So no, let's d prime exactly.

01:57 So for the configuration a, we have that the cross section of area is the length l times the height d for the configuration b, we have l time d prime.

02:17 But we can easily see that this d is two times and d, this d prime.

02:29 So it will be l times two times d, which is, we just take this, factorize we can easily see that the area of b is two times the area of a.

02:49 The next question of this problem asks which of the thickness is greater through which he flows in the arrangement in a or in the arrangement in b.

03:09 So this can easily easily be seen again.

03:14 We know that the thickness is this longitude in here because these are the parts through which the flow, the heat flows can go.

03:29 And this also is the thickness of this.

03:32 So from here we can see that if we call this, this must be two times c.

03:40 So if we call this, this must be two times c.

03:43 So if we call it, we call it.

03:45 The thickness of a as la, this is equal to 2 times c...

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