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 kg of the substance from −73 ° C to 27 ° C is nk, the value of n is _____. [Given: 0 K = −273 ° C.]

 The Quest for Heat: Unraveling Temperature-Dependent Specific Heat

The Quest for Heat: Unraveling Temperature-Dependent Specific Heat

Question:
The specific heat capacity of a mysterious substance defies convention—it changes with temperature! Governed by the formula $C=kT$, where $k$ is a constant and $T$ is absolute temperature, this substance challenges scientists to calculate the heat required to warm 1 kg of it from $-73^{\circ }\text{C}$ to $27^{\circ }\text{C}$. If the answer is $nk$, what is $n$?

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A Historical Prelude

The concept of specific heat dates back to the 18th century, with Joseph Black’s experiments on latent heat. But temperature-dependent specific heat? That’s a modern twist! Such complexities arise in materials like metals at cryogenic temperatures or gases undergoing phase changes. Today, we tackle this puzzle with calculus—a tool Black never had!

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Decoding the Problem

1. Key Insight: Unlike constant specific heat ($Q=mc\mathrm{\Delta }T$), here $C=kT$ varies. Integration is essential.

2. Units Matter: Temperatures must convert to Kelvin ($0K=-273^{\circ }\text{C}$):

   • $-73^{\circ }\text{C}=200K$

   • $27^{\circ }\text{C}=300K$

3. Integration: Compute $Q={\int }_{T_1}^{T_2}CdT$.

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Step-by-Step Solution

1. Formula Setup:

   $$Q=m{\int }_{T_1}^{T_2}kTdT(m=1\text{kg})$$

2. Simplify:

   $$Q=k{\int }_{200}^{300}TdT$$

3. Integrate:

   $$\int TdT=\frac{1}{2}{T}^2{\mid }_{200}^{300}=\frac{1}{2}(300^2-200^2)$$$$=\frac{1}{2}(90,\text{\hspace{0.17em}⁣}000-40,\text{\hspace{0.17em}⁣}000)=25,\text{\hspace{0.17em}⁣}000$$

4. Final Result:

   $$Q=25,\text{\hspace{0.17em}⁣}000k⟹n=25,\text{\hspace{0.17em}⁣}000$$

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Verification via Average Specific Heat

For linear $C(T)$, average $C$ over $200K$ to $300K$:

$$C_{\text{avg}}=\frac{k(200)+k(300)}{2}=250k$$

Heat using average:

$$Q=m\cdot C_{\text{avg}}\cdot \mathrm{\Delta }T=1\cdot 250k\cdot 100=25,\text{\hspace{0.17em}⁣}000k$$

Same result! Cross-check confirms $n=25,\text{\hspace{0.17em}⁣}000$.

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Real-World Analogy

Imagine heating a block of aluminum. At low temps, its specific heat rises linearly with $T$. From $-73^{\circ }\text{C}$ (Antarctic winter) to $27^{\circ }\text{C}$ (room temp), the energy required isn’t just proportional to $\mathrm{\Delta }T$—it’s a dance of calculus and physics!

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Answer
The value of $n$ is $\boxed{{\displaystyle 25000}}$.

Final Note: This problem bridges 18th-century thermodynamics with modern calculus—a testament to how science evolves. Next time you sip hot coffee, remember: even heating has hidden mathematical beauty! ☕✨