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

Answer 1

Correct Answer: 25000

To solve the problem, we need to find the heat required to raise the temperature of 1 kg of a substance from $-73^{\circ}C$ to $27^{\circ}C$.

1. Understanding the Specific Heat Capacity Relation:
The specific heat capacity of the substance is given by $C = kT$, where $k$ is a constant and $T$ is the absolute temperature.

2. Converting Temperatures to Kelvin:
We have $T_1 = -73^{\circ}C = -73 + 273 = 200 \, \text{K}$ and $T_2 = 27^{\circ}C = 27 + 273 = 300 \, \text{K}$.

3. Formula for Heat Required:
The heat required is given by:

$ Q = \int_{T_1}^{T_2} mC \, dT = \int_{T_1}^{T_2} m(kT) \, dT$
where $m = 1$ kg.

4. Simplifying the Integral:
We now have:

$ Q = \int_{200}^{300} 1(kT) \, dT = k \int_{200}^{300} T \, dT = k \left[ \frac{T^2}{2} \right]_{200}^{300}$

5. Calculating the Integral:
This becomes:

$ Q = k \left(\frac{300^2}{2} - \frac{200^2}{2} \right) = k \left(\frac{90000}{2} - \frac{40000}{2} \right)$

6. Final Simplification:
We get:

$ Q = k \left(\frac{50000}{2} \right) = k (25000)$

7. Solving for $n$:
Given that $Q = nk$, we have:

$ nk = 25000k$
This implies:

$ n = 25000$

Final Answer:
The final answer is $\boxed{25000}$.

Answer 2

To solve the problem, we calculate the heat required to raise the temperature of 1 kg of the substance when its specific heat capacity depends on temperature.

Given:
Specific heat capacity: \[ C = k T \] where \(k\) is a constant and \(T\) is the absolute temperature (in Kelvin).
Temperature change from \(-73^\circ C\) to \(27^\circ C\).

Step 1: Convert temperatures to Kelvin:
\[ T_1 = -73 + 273 = 200\, K \] \[ T_2 = 27 + 273 = 300\, K \]

Step 2: Calculate heat required:
Heat for mass \(m = 1\, kg\) with variable heat capacity: \[ Q = m \int_{T_1}^{T_2} C \, dT = \int_{200}^{300} k T \, dT = k \int_{200}^{300} T \, dT \] \[ Q = k \left[ \frac{T^2}{2} \right]_{200}^{300} = k \left( \frac{300^2}{2} - \frac{200^2}{2} \right) = \frac{k}{2} (90000 - 40000) = \frac{k}{2} \times 50000 = 25000 k \]

Step 3: Express heat as \(n k\):
\[ Q = n k \implies n = 25000 \]

Final Answer:
\[ \boxed{25000} \]