Conduction Heat Transfer in Thermodynamic Processes

Hello! In our last lesson, we focused on how mechanical work done on a system, through processes like compression or friction, gets converted into internal energy. This helped us understand the W term in the First Law, $Δ U = Q - W$ , from a new perspective.

Today, we shift our focus to the Q term. We will explore scenarios where the heat transferred to or from a thermodynamic system isn't a given value but is governed by the process of heat conduction through the system's boundaries. This is a classic integrated problem setup frequently seen in the JEE Advanced, where you need to combine principles from two different chapters: thermodynamics and heat transfer.

Our learning goal is to analyze thermodynamic processes where heat is exchanged via conduction through the system's boundaries. We will learn how to set up the energy balance equations that link the rate of heat flow to the change in the state of a gas.

1. Revisiting Heat Conduction: Fourier's Law

Before we can link conduction to thermodynamics, let's ensure we have a solid grasp of the governing principle of heat conduction: Fourier's Law. You've likely encountered this before, but a rigorous understanding is crucial for the complex problems ahead.

First, let's watch a video that provides an excellent conceptual and mathematical foundation for heat transfer by conduction.

Understanding Conduction and the Heat Equation

This video from 'The Efficient Engineer' offers a clear explanation of thermal conduction, starting from the molecular level and building up to Fourier's Law. It will give you a strong intuition for the physics involved.

Please watch the first 6 minutes and 14 seconds of the video. Pay close attention to: The definition of heat transfer rate, q (which we often denote as H or dQ/dt). The formulation of Fourier's Law: q = -kA(dT/dx). The concept of thermal conductivity, k, as a material property. How the law simplifies for a 1D steady-state case (heat flow through a wall or rod).

As the video explained, for a uniform rod of length $L$ and cross-sectional area $A$ , with its ends maintained at constant temperatures $T_{H}$ and $T_{C}$ , the rate of heat flow (or heat current) in the steady state is:

H = \frac{d Q}{d t} = \frac{k A ( T _{H} - T _{C} )}{L}

This equation is central to everything we'll do today.

2. The Electrical Analogy: Thermal Resistance

Your background in CS and mathematics makes you comfortable with powerful abstractions. One of the most useful in heat transfer is the analogy between heat flow and electrical circuits.

If we rewrite Fourier's Law as $H = \frac{T _{H} - T _{C}}{L / ( k A )}$ , it looks just like Ohm's Law, $I = \frac{Δ V}{R}$ .

Heat Current ( $H$ ) is analogous to Electric Current ( $I$ ).
Temperature Difference ( $Δ T$ ) is analogous to Potential Difference ( $Δ V$ ).
Thermal Resistance ( $R_{t h} = \frac{L}{k A}$ ) is analogous to Electrical Resistance ( $R$ ).

This analogy is incredibly powerful for solving problems with composite materials. To see how it's applied in a JEE context, let's review a dedicated resource.

Understanding Fourier's Law of Heat Conduction

This article from Allen Career Institute is tailored for JEE preparation. It concisely summarizes Fourier's Law and the extremely useful formulas for series and parallel combinations of thermal resistances.

Please read through sections 1 to 8 of the article. Focus on: The formula for thermal resistance (Section 5). The formulas for equivalent thermal conductivity and resistance for series and parallel combinations (Section 7). The method for finding the junction temperature in a series combination (Section 8).

The key takeaways from this are:

Series Combination: Rods are connected end-to-end. Heat current is the same through all. $R_{e q} = R_{1} + R_{2} + ...$
Parallel Combination: Rods are connected side-by-side between the same two temperatures. Total heat current is the sum of currents. $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + ...$

Let's watch a solved JEE Advanced problem that uses these exact concepts.

JEE Advanced Physics 2013 Paper 1 #5 (#9): Heat Conductivity

This video by Michel van Biezen tackles a classic JEE Advanced problem comparing heat flow through series and parallel configurations. It's a perfect application of the thermal resistance concept.

Watch the entire video. Notice how he calculates the heat flow rate (dq/dt) for each configuration using the thermal resistance formulas you just reviewed. The core of the problem lies in setting up the ratio of these rates.

3. Bridging Conduction and Thermodynamics

Now for the main event: connecting these ideas. We will analyze a system where the heat $Q$ flowing via conduction causes a change in the internal energy $U$ of a gas.

Consider the setup shown in the problem below: two thermally insulated containers with gases are connected by a conducting rod.

JEE Advanced Problem: Heat Conduction Between Gas Containers — A JEE Advanced problem showing two containers, one with O₂ at temperature T₁ and another with He at T₂, connected by a solid cylindrical rod. Heat transfer occurs only through the rod.

Let's assume the gases are kept at constant volume. The temperature of the hot gas ( $T_{1}$ ) will decrease, and the temperature of the cold gas ( $T_{2}$ ) will increase, until they reach a common final temperature.

The heat current from the hot to the cold container at any instant is:
$H = \frac{d Q}{d t} = \frac{k A ( T _{1} - T _{2} )}{L}$

Now, let's apply the First Law of Thermodynamics to each gas. Since volume is constant, work done $W = 0$ .

For the hot gas (Container A): It loses heat. The heat transfer from the gas is $d Q_{1}$ . So, the change in its internal energy is $d U_{1} = - d Q_{1}$ .
$n_{1} C_{V 1} d T_{1} = - d Q_{1} ⟹ \frac{d Q _{1}}{d t} = - n_{1} C_{V 1} \frac{d T _{1}}{d t}$
The heat leaving gas 1 is the heat flowing through the rod, so $\frac{d Q _{1}}{d t} = H$ .
$\frac{k A ( T _{1} - T _{2} )}{L} = - n_{1} C_{V 1} \frac{d T _{1}}{d t} (Eq. 1)$
For the cold gas (Container B): It gains heat $d Q_{2}$ . Its internal energy change is $d U_{2} = d Q_{2}$ .
$n_{2} C_{V 2} d T_{2} = d Q_{2} ⟹ \frac{d Q _{2}}{d t} = n_{2} C_{V 2} \frac{d T _{2}}{d t}$
The heat entering gas 2 is also $H$ .
$\frac{k A ( T _{1} - T _{2} )}{L} = n_{2} C_{V 2} \frac{d T _{2}}{d t} (Eq. 2)$

The problem in the image asks for the time it takes for the temperature difference $Δ T = T_{1} - T_{2}$ to change. Let's find an expression for $\frac{d ( Δ T )}{d t}$ .

From Eq. 1: $\frac{d T _{1}}{d t} = - \frac{k A}{L n _{1} C _{V 1}} (T_{1} - T_{2})$
From Eq. 2: $\frac{d T _{2}}{d t} = \frac{k A}{L n _{2} C _{V 2}} (T_{1} - T_{2})$

Now, $\frac{d ( Δ T )}{d t} = \frac{d T _{1}}{d t} - \frac{d T _{2}}{d t}$ . Substituting the above expressions:
$\frac{d ( Δ T )}{d t} = - (\frac{k A}{L n _{1} C _{V 1}} + \frac{k A}{L n _{2} C _{V 2}}) (T_{1} - T_{2})$
This is a first-order linear differential equation. If we let the constant term in the parenthesis be $C$ , we have:
$\frac{d ( Δ T )}{d t} = - C (Δ T) ⟹ \frac{d ( Δ T )}{Δ T} = - C d t$

Integrating this from an initial temperature difference $Δ T_{i}$ at $t = 0$ to a final difference $Δ T_{f}$ at time $t$ gives:
$\int_{Δ T_{i}}^{Δ T_{f}} \frac{d ( Δ T )}{Δ T} = \int_{0}^{t} - C d t$
$ln (\frac{Δ T _{f}}{Δ T _{i}}) = - C t ⟹ t = \frac{1}{C} ln (\frac{Δ T _{i}}{Δ T _{f}})$

This is a standard result for relaxation processes and a powerful technique for JEE Advanced.

Test your understanding!

Two conducting cylinders of equal length $L$ but different radii ( $r_{2} = 2 r_{1}$ ) are connected in series between two heat baths at $T_{1} = 300$ K and $T_{2} = 100$ K. The thermal conductivities are $K_{1}$ and $K_{2}$ . In steady state, the temperature at the junction is found to be 200 K. Find the ratio $K_{1} / K_{2}$ . (This is problem 23 from the Allen PDF).

Show answer

In steady state, the heat current $H$ must be the same through both cylinders. Let $T_{j} = 200$ K be the junction temperature.

Heat current through the first cylinder:
$H_{1} = \frac{K _{1} A _{1} ( T _{1} - T _{j} )}{L} = \frac{K _{1} ( π r _{1}^{2} ) ( 300 - 200 )}{L} = \frac{100 K _{1} π r _{1}^{2}}{L}$

Heat current through the second cylinder:
$H_{2} = \frac{K _{2} A _{2} ( T _{j} - T _{2} )}{L} = \frac{K _{2} ( π r _{2}^{2} ) ( 200 - 100 )}{L} = \frac{K _{2} ( π ( 2 r _{1} ) ^{2} ) ( 100 )}{L} = \frac{400 K _{2} π r _{1}^{2}}{L}$

Since $H_{1} = H_{2}$ :
$\frac{100 K _{1} π r _{1}^{2}}{L} = \frac{400 K _{2} π r _{1}^{2}}{L}$
Canceling the common terms ( $100, π, r_{1}^{2}, L$ ):
$K_{1} = 4 K_{2}$
$\frac{K _{1}}{K _{2}} = 4$

4. Practice Problem

Let's try another integrated problem that combines conduction with a topic from a previous module, thermal expansion.

[PDF] JEE Advanced Physics 10 Years Topicwise Questions with Solutions

This problem from the JEE Advanced question bank requires you to first find the temperature at the junction of two wires and then use that to calculate thermal expansion.

Find and attempt Question 34 from the PDF. The setup involves two wires, PQ and RS, of different thermal conductivities soldered together. Follow these steps: Identify the configuration (series or parallel). Set the heat currents equal to find the temperature of the junction (Q and R). Use the temperature of wire PQ to calculate its change in length due to thermal expansion.

Let's outline the solution strategy.

Find Junction Temperature $T_{j}$ : The wires are in series. In steady state, $H_{P Q} = H_{R S}$ .
$\frac{K _{P Q} A ( T _{j} - T _{P} )}{L _{P Q}} = \frac{K _{R S} A ( T _{S} - T _{j} )}{L _{R S}}$
Given $K_{P Q} = 2 K_{R S}$ , $L_{P Q} = L_{R S} = 1$ m, $T_{P} = 1 0^{\circ}$ C, and $T_{S} = 40 0^{\circ}$ C.
$2 K_{R S} (T_{j} - 10) = K_{R S} (400 - T_{j})$
$2 T_{j} - 20 = 400 - T_{j} ⟹ 3 T_{j} = 420 ⟹ T_{j} = 14 0^{\circ} C$
Calculate Thermal Expansion of PQ: The change in length is $Δ L = α L_{0} Δ T$ .
The average temperature of wire PQ is not needed here; the temperature gradient is linear. The change in length depends on the average temperature, which for a linear profile is $(T_{P} + T_{j}) /2$ . However, the formula $Δ L = α L_{0} Δ T_{a v g}$ works. Wait, a simpler approach for linear temperature distribution is just using the total temperature change across its ends for the formula of expansion of the center of mass, but let's stick to the JEE convention.
The problem is slightly ambiguous. In a strict sense, each segment $d x$ of the rod expands based on its local temperature $T (x)$ . $Δ L = \int_{0}^{L} α T (x) d x - α L T_{ini t ia l}$ .
For a linear profile $T (x) = T_{P} + \frac{T _{j} - T _{P}}{L} x$ , the integral gives $Δ L = α L (\frac{T _{P} + T _{j}}{2} - T_{ini t ia l})$ .
So, $Δ T_{a v g} = \frac{10 + 140}{2} - 10 = 75 - 10 = 6 5^{\circ}$ C.
Let's re-read the problem carefully... "change in length of the wire PQ". It seems we need to find the total elongation of PQ. The temperature of P is 10°C, and Q is 140°C. Initial temperature was 10°C.
Average temperature of PQ is $(10 + 140) /2 = 7 5^{\circ}$ C. The average temperature rise is $75 - 10 = 6 5^{\circ}$ C.
$Δ L_{P Q} = α_{P Q} L_{P Q} (Δ T_{a v g}) = (1.2 \times 1 0^{- 5}) \times (1) \times (65) = 78 \times 1 0^{- 5} m = 0.78 mm$

Conclusion

Today, we built a crucial bridge between two major topics in thermal physics. By connecting Fourier's Law of conduction with the First Law of Thermodynamics, we unlocked a powerful method for analyzing how systems evolve over time when their boundaries conduct heat.

Key Takeaways:

Heat transfer via conduction is quantified by Fourier's Law, with the rate of heat flow given by $H = d Q / d t = - k A (d T / d x)$ .
The concept of thermal resistance ( $R_{t h} = L / k A$ ) and its analogy to electrical resistance provides a powerful problem-solving framework, especially for composite conductors in series and parallel.
In integrated problems, the heat current $H$ is the rate of heat transfer $d Q / d t$ that drives changes in a system's internal energy, described by $d U / d t = n C_{V} (d T / d t)$ .
This connection often results in a first-order differential equation, leading to an exponential change in temperature over time, a pattern you should be prepared to analyze.

Preview of the Next Lesson:
In our next lesson, we'll add another essential technique to your problem-solving arsenal: applying the binomial approximation in thermal expansion and thermodynamic calculations for small changes. This mathematical shortcut is indispensable for simplifying expressions and quickly solving problems involving small fractional changes, a common feature in JEE Advanced questions.

Recommended resources

1. Revisiting Heat Conduction: Fourier's Law

2. The Electrical Analogy: Thermal Resistance

3. Bridging Conduction and Thermodynamics

4. Practice Problem

Conclusion