Heat Transfer with Temperature-Dependent Specific Heat

Hello! Welcome back to our study of thermodynamics.

In our last lesson, we established the principle of calorimetry for mixing problems, using the formula $Q = m c Δ T$ to calculate heat exchange. This formula works perfectly when the specific heat capacity, $c$ , is constant. Today, we'll address the learning outcome of solving problems where the specific heat capacity itself varies with temperature.

This scenario is common in real materials, especially at very low temperatures, and tackling it requires a more powerful mathematical tool you're already very familiar with: calculus. We will move from simple multiplication to integration to find the total heat exchanged. This is a classic step-up in complexity found in JEE Advanced problems.

1. From an Approximation to an Exact Calculation

The formula $Q = m c Δ T$ is an excellent approximation, but it's built on the assumption that $c$ is constant across the temperature change $Δ T$ . What if it's not? What if the specific heat capacity is a function of temperature, $c (T)$ ?

In that case, we can only consider the formula to be accurate for an infinitesimally small temperature change, $d T$ . The small amount of heat, $d Q$ , required to produce this change is:
$d Q = m c (T) d T$

To find the total heat $Q$ needed to change the temperature from an initial temperature $T_{1}$ to a final temperature $T_{2}$ , we must sum all these infinitesimal contributions. This summation is precisely what a definite integral does.
$Q = \int_{T_{1}}^{T_{2}} m c (T) d T$

This integral represents the total heat transferred by accumulating the small amounts of heat over the entire temperature path.

To see a formal treatment of this concept, let's turn to a resource.

1.5: Heat Transfer, Specific Heat, and Calorimetry - Physics LibreTexts

The following text from Physics LibreTexts formally introduces the concept of temperature-dependent heat capacity. It explains the transition from the algebraic formula to the integral form and provides a relevant example.

Please read the subsection titled 'Temperature-Dependent Heat Capacity'. It demonstrates how the differential heat dQ leads to the integral form Q = m ∫c(T)dT. Pay attention to the example where the specific heat of salt (NaCl) is proportional to T³, a model that is important in solid-state physics.

2. A Worked Example: Variable Specific Heat

Now that we have the theory, let's watch a concise video that applies this integral method to a typical physics problem.

18. Advance Illustration | Thermal Expansion & Calorimetry | Substance with Specific Heat

This video from Physics Galaxy solves a problem where the specific heat varies with the cube of the absolute temperature, very similar to the Debye T³ law mentioned in the text you just read.

Watch the entire video (it's under 3 minutes). Observe how the instructor first defines the differential heat element dQ using the given specific heat function and then integrates it over the specified temperature range to find the total heat.

The process is straightforward:

Write the expression for infinitesimal heat, $d Q = m c (T) d T$ .
Set up the definite integral for the total heat, $Q$ , with the initial and final temperatures as the limits of integration.
Solve the integral to find the answer.

3. JEE Advanced Style Problem

Let's now apply this to a problem structured in the way you might see on the JEE Advanced exam. Pay close attention to the details, especially the units and temperature scales.

Test your understanding!

The specific heat capacity of a substance is temperature dependent and is given by the formula $c = k T$ , where $k$ is a constant and $T$ is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from −73 °C to 27 °C is $nk$ , what is the value of $n$ ?

Before checking the answer, try to solve it yourself. A crucial first step in many thermodynamics problems is converting all temperatures to the absolute scale (Kelvin).

Show answer

This problem is a direct application of the integration method we've just discussed. The solution can be found in the collegedunia.com resource, but let's walk through it step-by-step.

Step 1: Convert Temperatures to Kelvin
The formula $c = k T$ uses absolute temperature $T$ . Therefore, we must convert the given Celsius temperatures to Kelvin.

Initial Temperature: $T_{1} = - 73° C + 273 = 200 K$
Final Temperature: $T_{2} = 27° C + 273 = 300 K$

Step 2: Set up the Integral for Total Heat (Q)
We use the integral formula for variable specific heat:
$Q = \int_{T_{1}}^{T_{2}} m c (T) d T$
Substitute the given values: $m = 1$ kg and $c (T) = k T$ .
$Q = \int_{200}^{300} (1) \cdot (k T) d T$

Step 3: Solve the Integral
Since $k$ is a constant, we can pull it out of the integral.
$Q = k \int_{200}^{300} T d T$
The integral of $T$ is $\frac{T ^{2}}{2}$ . We evaluate this from 200 K to 300 K.
$Q = k [\frac{T ^{2}}{2}]_{200}^{300}$
$Q = k (\frac{30 0 ^{2}}{2} - \frac{20 0 ^{2}}{2})$

Step 4: Calculate the Final Value
$Q = \frac{k}{2} (90000 - 40000) = \frac{k}{2} (50000) = 25000 k$

Step 5: Find the value of n
The problem states that the required heat is $nk$ . We found the heat to be $25000 k$ .
$nk = 25000 k$
Therefore, $n = 25000$ .

This problem highlights a common pitfall: failing to convert to Kelvin when the physical law is expressed in terms of absolute temperature.

4. A Note on Heat Capacity vs. Specific Heat Capacity

It's important to be precise with terminology:

Specific Heat Capacity ( $c$ ): The heat required to raise the temperature of a unit mass of a substance by one degree. Units are typically J/(kg·K) or cal/(g·°C).
Heat Capacity ( $C$ ): The heat required to raise the temperature of an entire object by one degree. It is the product of the object's mass and its specific heat capacity ( $C = m c$ ). Units are J/K or cal/°C.

If a problem gives you a temperature-dependent heat capacity $C (T)$ , the formula for heat becomes slightly simpler, as mass is already included:
$Q = \int_{T_{1}}^{T_{2}} C (T) d T$
Always read the problem carefully to determine whether you are given $c (T)$ or $C (T)$ .

5. Advanced Application: Mixing Problems with Variable Specific Heat

How does this concept integrate with the mixing problems from our last lesson? Imagine mixing a substance 'A' with a constant specific heat $c_{A}$ and a substance 'B' with a variable specific heat $c_{B} (T)$ .

Let's say hot substance 'A' (at $T_{A}$ ) is mixed with cold substance 'B' (at $T_{B}$ ). They reach a final equilibrium temperature $T_{f}$ . The principle of calorimetry (Heat Lost = Heat Gained) still holds:

Heat Lost by A: $Q_{lost} = m_{A} c_{A} (T_{A} - T_{f})$ (Assuming $c_{A}$ is constant)
Heat Gained by B: $Q_{gained} = \int_{T_{B}}^{T_{f}} m_{B} c_{B} (T) d T$

The energy balance equation would be:
$m_{A} c_{A} (T_{A} - T_{f}) = m_{B} \int_{T_{B}}^{T_{f}} c_{B} (T) d T$

Solving this equation for the final temperature $T_{f}$ can become algebraically complex, as $T_{f}$ appears both outside and inside the integral's limit. This is the kind of multi-concept integration that JEE Advanced problems love to test. You don't need to solve such a problem right now, but it's crucial to understand how the principles combine.

Conclusion

In this lesson, we've extended our understanding of calorimetry to handle more realistic scenarios where specific heat is not constant. Your strong foundation in calculus makes this a natural extension of the basic principles.

Key Takeaways:

When specific heat capacity $c$ is a function of temperature $T$ , the simple formula $Q = m c Δ T$ is replaced by the integral form: $Q = \int m c (T) d T$ .
Solving these problems requires careful setup of the integral with the correct limits (initial and final temperatures).
Always be vigilant about units, especially ensuring that temperatures are converted to Kelvin if the specific heat formula depends on absolute temperature.

In our next lesson, we will introduce another crucial concept: phase changes. We will explore what happens when heat added to a substance doesn't change its temperature but instead causes it to melt or boil. This will introduce the idea of latent heat, adding another essential tool to our problem-solving arsenal.