A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The me

Q: Why do we use $P = C \frac{dT}{dt}$ instead of $Q = mc\Delta T$?

$Q = mc\Delta T$ is for a constant heat capacity over a finite temperature change. Here, the temperature is changing continuously, and the heat capacity itself might depend on time or temperature. So, we must use the differential form: rate of heat transfer (Power) equals heat capacity times the rate of change of temperature.

Q: What is the difference between specific heat and heat capacity here?

Specific heat ($s$ or $c$) is per unit mass ($C = ms$). The question asks for the 'heat capacity of the metal' as a whole, which is the extensive property $C$. We don't need the mass to solve this.

Q: Why did we have to eliminate $t$ at the end?

Look at the options! They are all expressed in terms of $T(t)$ and constants. In physics problems, it's standard practice to express material properties (like heat capacity) as a function of state variables (like temperature) rather than time.

Q: Is it physically realistic for heat capacity to depend on temperature?

Absolutely! In fact, at very low temperatures, the heat capacity of metals follows the Debye $T^3$ law. At higher temperatures, it might have other non-linear dependencies. This problem models a specific non-linear behavior.

Q: I got confused with the algebra when substituting $t$. Any tips?

Instead of solving for $t$ completely (which would involve a 4th power), notice that your expression for $C$ contains $t^{3/4}$, which is just $(t^{1/4})^3$. Since the original equation easily gives you $t^{1/4}$, substitute that directly. It saves time and reduces errors!

Question

A current carrying wire heats a metal rod. The wire provides a constant power ($P$) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature ($T$) in the metal rod changes with time ($t$) as :

$T(t) = T_0 (1 + \beta t^{1/4})$

where $\beta$ is a constant with appropriate dimension while $T_0$ is a constant with dimension of temperature. The heat capacity of the metal is :

Options:

(A) $\frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$

(B) $\frac{4P(T(t) - T_0)}{\beta^4 T_0^2}$

(C) $\frac{4P(T(t) - T_0)^4}{\beta^4 T_0^5}$

(D) $\frac{4P(T(t) - T_0)^2}{\beta^4 T_0^3}$

Sigma Percentile · Accepted Answer

### The Calculus of Calorimetry: Unraveling a Non-Linear Heating Process

Imagine you are conducting an experiment in a perfectly insulated laboratory. You have a metal rod, and you wrap a heating wire around it. You turn on the power supply, delivering a constant electrical power $P$ to the rod. Because the container is perfectly insulated, not a single joule of heat escapes to the surroundings. All the electrical energy is converted into internal thermal energy, raising the temperature of the rod.

However, when you plot the temperature $T$ against time $t$, you notice something peculiar. The temperature doesn't rise in a straight line. Instead, it follows a curve given by:

$$T(t) = T_0 (1 + \beta t^{1/4})$$

The fact that the temperature grows proportionally to $t^{1/4}$ means that as time goes on, the rate at which the rod heats up slows down significantly. Why would a constant power supply result in a slowing temperature rise? The physical implication is profound: **the heat capacity of the metal is not constant; it is increasing as the rod gets hotter!** Let's use the power of calculus to find the exact mathematical form of this heat capacity.

### The Master Equation of Heating

From the fundamental principles of calorimetry, the rate at which heat $Q$ is supplied to an object is equal to its heat capacity $C$ multiplied by the rate of change of its temperature. Since the rate of heat supply is simply the electrical power $P$, we can write our master differential equation:

$$P = \frac{dQ}{dt} = C \frac{dT}{dt}$$

Our goal is to find $C$. To do this, we need to extract $\frac{dT}{dt}$ from the given temperature function.

### Differentiating the Temperature Profile

Let's take the derivative of the temperature function with respect to time $t$.

$$T(t) = T_0 + T_0 \beta t^{1/4}$$

Applying the power rule of differentiation, the constant $T_0$ vanishes, and the exponent $\frac{1}{4}$ comes down as a multiplier:

$$\frac{dT}{dt} = T_0 \beta \left( \frac{1}{4} t^{-3/4} \right) = \frac{\beta T_0}{4} t^{-3/4}$$

### Isolating the Heat Capacity

Now, we substitute this derivative back into our master equation:

$$P = C \left( \frac{\beta T_0}{4} t^{-3/4} \right)$$

Rearranging this to solve for the heat capacity $C$, we get:

$$C = \frac{4P}{\beta T_0} t^{3/4}$$

We have successfully found the heat capacity as a function of time. However, if you look at the options provided in the question, none of them contain the variable $t$. They are all expressed in terms of the instantaneous temperature $T(t)$. This is a common practice in physics: material properties are usually expressed as functions of state variables (like temperature) rather than time.

### The Final Substitution

To eliminate $t$, we must return to our original temperature equation and isolate the time variable.

$$T(t) - T_0 = T_0 \beta t^{1/4}$$

Dividing by $T_0 \beta$, we isolate $t^{1/4}$:

$$t^{1/4} = \frac{T(t) - T_0}{\beta T_0}$$

Our expression for heat capacity contains $t^{3/4}$, which is simply the cube of $t^{1/4}$. Therefore, we can substitute our isolated expression directly:

$$C = \frac{4P}{\beta T_0} \left( t^{1/4} \right)^3$$

$$C = \frac{4P}{\beta T_0} \left( \frac{T(t) - T_0}{\beta T_0} \right)^3$$

Expanding the cube in the denominator gives us our final, elegant result:

$$C = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$$

This result beautifully confirms our initial physical intuition. As the temperature $T(t)$ increases, the term $(T(t) - T_0)^3$ grows rapidly, meaning the heat capacity $C$ becomes very large. This is exactly why a constant power supply struggles to raise the temperature quickly at later times!

Select Answer:

Visualized Solution

$P = \frac{d Q}{d t}$

$P = C \frac{d T}{d t}$

$T (t) = T_{0} (1 + β t^{1/4})$

$\frac{d T}{d t} = \frac{β T _{0}}{4} t^{- 3/4}$

$C = \frac{4 P}{β T _{0}} t^{3/4}$

$t^{1/4} = \frac{T ( t ) - T _{0}}{β T _{0}}$

$C = \frac{4 P ( T ( t ) - T _{0} ) ^{3}}{β ^{4} T _{0}^{4}}$

$Conclusion$

The Sigma Insight: Calorimetry

The Calculus of Calorimetry

The Master Equation of Heating

Differentiating the Temperature Profile

Isolating the Heat Capacity

The Final Substitution

Similar Questions

Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is

A substance of mass $M kg$ requires a power input of $P watts$ to remain in the molten state at its melting point. When the power source is turned off, the sample completely solidifies in time $t seconds$ . The latent heat of fusion of the substance is …… .

Time taken by a $836 W$ heater to heat $1 L$ of water from $1 0^{\circ} C$ to $4 0^{\circ} C$ is

Water of volume 2 L in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at a rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C? [Specific heat of water is 4.2 kJ/kg]

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=dtdQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=CdtdT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T(t)=T0​(1+βt1/4)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdT​=4βT0​​t−3/4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=βT0​4P​t3/4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1/4=βT0​T(t)−T0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=β4T04​4P(T(t)−T0​)3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion