Two rods, one made of brass and the other made of copper, are joined end to end. The length of t

Two rods, one made of brass and the other made of copper,...

Key Concepts

Heat Conduction

Heat conduction is the process by which thermal energy is transferred through materials due to a temperature gradient. This concept is fundamental when analyzing how energy moves from a hot region (boiling water) through the rod to a cold region (ice-water mixture), and it is inherently governed by the intrinsic thermal properties of the materials involved.

Fourier's Law

Fourier's Law of heat conduction states that the rate of heat transfer through a material is proportional to the negative temperature gradient and the material's thermal conductivity, and is given by the equation Q/t = kA (?T/L). This law is crucial for quantifying the amount of energy that flows through each segment of a composite conductor.

Thermal Resistance in Series

When different materials are connected in series, such as the brass and copper sections of a composite rod, each segment acts as a thermal resistor. The thermal resistance depends on the material's properties, length, and cross-sectional area, and the total resistance is the sum of the individual resistances. This analogy helps in determining the temperature drops and the interface temperature between different materials.

Steady-State Temperature Distribution

In steady-state conditions, the temperature distribution in a system does not change with time, meaning that the heat transfer rate is constant along the conductor. This assumption is vital for equating the heat flow through the different materials in the composite rod and for calculating the temperature at the junction where they are joined.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy required to change a substance from a solid to a liquid without a change in temperature. In heat transfer problems involving phase changes, such as melting ice, this concept is applied to determine how much ice can be melted by the energy conducted through the system over a given period.

Transcript

00:01 Okay, so this question asks us to calculate the temperature between the joints or the point where a brass rod and a copper road are joined together.

00:13 All right.

00:13 And we told that in this question that we have the brass rod fixed inside boiling water.

00:19 And then we have the copper road, one of its end, also fixed in ice and water fixture.

00:26 All right.

00:26 So let me draw this diagram to represent the situation that was.

00:30 We've been asked to solve.

00:32 All right.

00:33 So this is my car.

00:35 Let's see my bull containing boiling water.

00:39 All right.

00:39 And then inside it's a brush rod.

00:43 All right.

00:44 And then let's see this is my copper word.

00:48 All right.

00:48 Which is also fixed inside ice and water mixture.

00:55 Ice and water mixture.

00:58 All right.

00:59 So this is it.

01:01 Eyes and water mixture.

01:03 All right.

01:03 Forgive my diagram.

01:05 Okay, all right.

01:06 So we told that the brass has a length of 0 .2 ,000 meters, and then the length of the copper is 0 .8 meters.

01:24 All right.

01:25 Now, we want to determine the temperature, the common temperature between the joints where they are joined together.

01:31 That is the point where we want to determine their temperatures.

01:33 All right.

01:35 Now, we need to understand that heat always flows from a hotter region to a cold region.

01:41 All right.

01:42 And we know that the brush rule, which is fixing a hotter water, will have heat flowing through it to the cold area inside where we can locate our copper.

01:52 All right.

01:54 So in this situation, we're going to talk about heat current.

01:57 All right.

01:57 Now, the equation talking about heat current is giving as the heat current.

02:04 Is given us the thermal conductivity times the area into bracket a changing temperature minus, let's say the hotter one, minus the cold one, divided by the length of the rod that we are considering.

02:18 Lent of the road that we are considering.

02:21 Now in this question, since we're talking about two roads, since we're talking about two roads, we're going to change up this equation a bit to suit the situation that we are studying.

02:33 All right.

02:34 Now, since there is a common temperature between the joints where this road are set up, then we need to change our equation to look like this.

02:41 K, let me say k subscript b, that stands for brass times the area to bracket t, subscript h minus.

02:51 Now, the common temperature we're going to represent with t, all right, divided by the length, the brass l subscript b.

02:58 And all of this is equal to k subscript c times a, into bracket t h.

03:06 Oh, sorry, forgive me, to bracket t minus tc.

03:11 Now, t minus tc because the t will have a higher temperature than the temperature of the cold area.

03:19 And all of this divided by the length of the copper.

03:22 All right.

03:24 Okay.

03:25 Now, we need to also understand this, that the joints where we have their temperatures, the same, what it means is that the heat current is at a point in both roads are the same.

03:34 So we can straight away and equate these equations.

03:38 We need to understand that.

03:40 All right.

03:41 So we have our length of the road already given.

03:46 Okay, i've already written that.

03:48 So our thermal conductivity for brass is giving us one zero nine.

03:53 We can look up that in both textbooks or search them online.

03:58 What per meter per kelvin? and then that of copper is straight.

04:04 8, 5, 0 was a meter, a kelvin.

04:13 All right, good.

04:14 Now let's go on straight away and put in these values.

04:18 Okay, before the let's comment the temperatures.