The specific heat capacity of a substance is temperature dependent and is given by the formula C = k

Q: Why can't I just use the formula $Q = mC\Delta T$?

That formula only works when the specific heat capacity $C$ is a constant value. In this problem, $C$ changes as the temperature changes ($C = kT$). Because it's continuously varying, you have to use integration to sum up the tiny amounts of heat $dQ$ required for each tiny change in temperature $dT$.

Q: Why do I have to convert the temperatures to Kelvin? Can't I just integrate from -73 to 27?

The problem explicitly states that $T$ in the formula $C = kT$ is the *absolute temperature*. Absolute temperature is measured in Kelvin. If you integrate using Celsius, you are mathematically treating $0^\circ\text{C}$ as zero specific heat, which is physically incorrect. You must shift the scale to Kelvin.

Q: What is the physical meaning of the area under the $C$ vs $T$ graph?

For a $1\text{ kg}$ mass, the area under the specific heat capacity ($C$) versus temperature ($T$) graph represents the total heat energy ($Q$) required to change the substance's temperature between those two points. This is exactly what the integral $\int C dT$ calculates.

Q: Does the mass matter in this problem?

Yes, it does! The general formula for heat is $dQ = m C dT$. In this specific problem, the mass $m$ is given as $1\text{ kg}$, which makes it mathematically invisible in the integral (since multiplying by 1 changes nothing). But conceptually, it is still there.

Q: What if the specific heat was given as $C = a + bT$ instead?

The process would be exactly the same! You would set up the integral $Q = \int (a + bT) dT$ and evaluate it using the power rule for integration, resulting in $Q = [aT + \frac{bT^2}{2}]$ evaluated from the initial to the final Kelvin temperatures.

Question

The specific heat capacity of a substance is temperature dependent and is given by the formula $C = kT$, where $k$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of $1	ext{ kg}$ of the substance from $-73^\circ	ext{C}$ to $27^\circ	ext{C}$ is $nk$, the value of $n$ is _____.

[Given: $0	ext{ K} = -273^\circ	ext{C}$]

Sigma Percentile · Accepted Answer

### The Trap of the Constant

When we first learn about heat transfer, the formula $Q = mC\Delta T$ is drilled into our heads. It's elegant, simple, and works perfectly—as long as the specific heat capacity $C$ is a constant. But nature isn't always so accommodating. In this problem, we are introduced to a substance where the specific heat capacity is temperature-dependent, given by $C = kT$.

This means that as the substance gets hotter, it becomes increasingly "stubborn" and requires more heat to raise its temperature by the same amount. If you try to plug this into the standard formula, you'll immediately hit a wall. Which temperature do you use? The initial? The final? The average? None of them will give you the exact answer. We need a more powerful tool.

### The Calculus of Heat

Enter calculus. When a quantity is continuously changing, we can't look at the macroscopic picture all at once. We have to zoom in. Imagine adding a microscopic amount of heat, $dQ$, which causes a microscopic rise in temperature, $dT$. Over this tiny interval, the temperature is essentially constant, so our trusty formula works in its differential form: $dQ = m C dT$.

To find the total heat $Q$, we must sum up all these microscopic heat contributions from our starting temperature to our ending temperature. This continuous summation is exactly what an integral does. Geometrically, if we plot $C$ against $T$, the total heat required is the area under the curve between our initial and final states.

### The Absolute Temperature Catch

Before we rush into integrating, there is a classic trap waiting for us: units. The problem gives us temperatures in Celsius ($-73^\circ	ext{C}$ and $27^\circ	ext{C}$), but the formula $C = kT$ explicitly states that $T$ is the **absolute temperature**.

Absolute temperature is measured in Kelvin. If we integrate using Celsius, our limits will be negative and positive, and the math will completely break down, giving us a physically meaningless result. We must convert our bounds:

$T_i = -73 + 273 = 200	ext{ K}$

$T_f = 27 + 273 = 300	ext{ K}$

### Executing the Math

Now we are ready to set up our integral. We know the mass $m = 1	ext{ kg}$ and $C = kT$.

$$Q = \int_{T_i}^{T_f} m C \, dT$$

Substituting our values:

$$Q = \int_{200}^{300} (1)(kT) \, dT$$

Since $k$ is a constant, we can pull it out of the integral:

$$Q = k \int_{200}^{300} T \, dT$$

The integral of $T$ with respect to $T$ is simply $\frac{T^2}{2}$. Now we evaluate this from $200$ to $300$:

$$Q = k \left[ \frac{T^2}{2} ight]_{200}^{300}$$

$$Q = \frac{k}{2} \left( 300^2 - 200^2 ight)$$

Let's do the arithmetic. $300^2$ is $90,000$, and $200^2$ is $40,000$.

$$Q = \frac{k}{2} (90000 - 40000)$$

$$Q = \frac{k}{2} (50000)$$

$$Q = 25000k$$

The problem states that the total heat required is $nk$. By comparing our result to this expression, it is crystal clear that **$n = 25000$**.

This problem is a beautiful reminder that physics is not just about memorizing formulas; it's about understanding the underlying principles. When the rules of the game change—like a variable specific heat—calculus provides the framework to adapt and conquer.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Process}

\text{Temperature Conversion}

\text{Heat Equation for Variable Specific Heat}

\text{Setting up the Integral}

\text{Evaluating the Integral}

\text{Applying Limits}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Calorimetry

The Trap of the Constant

The Calculus of Heat

The Absolute Temperature Catch

Executing the Math

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