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 $\boxed{25000}$.
Three conductors of same length having thermal conductivity \(k_1\), \(k_2\), and \(k_3\) are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is ________ °C. (Given: \(k_1\) = 60 Js⁻¹m⁻¹K⁻¹,\(k_2\) = 120 Js⁻¹m⁻¹K⁻¹, \(k_3\) = 135 Js⁻¹m⁻¹K⁻¹)