A cylindrical conductor has a resistance R. When the conductor is at a temperature `(T)` above its surrounding temperature `(T_0)`, the ratio of thermal power dissipated by the conductor to its excess temperature `(DeltaT = T – T_(0))` above surrounding is a known constant `k`. The conductor is connected to a cell of emf `V`. Initially, the conductor was at room temperature `T_(0)`. Mass and specific heat capacity of the conductor are m and s respectively.
(i) find the time (t) dependence of the temperature (T) of the conductor after it is connected to the cell. Assume no change in resistance due to temperature.
(ii) find the temperature of the conductor after a long time.

Answer

To solve the problem, we will break it down into two parts as specified in the question.

### Part (i): Find the time (t) dependence of the temperature (T) of the conductor after it is connected to the cell.

1. **Understanding the Power Dissipation**:
The power (P) dissipated in the resistor (conductor) when connected to a cell of emf V is given by:
\[ P = \frac{V^2}{R} \]

2. **Heat Transfer to Surroundings**:
The thermal power dissipated by the conductor to its surroundings is proportional to the excess temperature \((\Delta T = T - T_0)\). According to the problem, this can be expressed as:
\[ P_{\text{dissipated}} = k \Delta T = k (T - T_0) \]

3. **Setting Up the Equation**:
The power dissipated in the conductor equals the rate of heat dissipated to the surroundings:
\[ \frac{V^2}{R} = k (T - T_0) \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ T - T_0 = \frac{V^2}{kR} \]
Thus,
\[ T = T_0 + \frac{V^2}{kR} \]

5. **Heat Capacity and Temperature Change**:
The change in temperature of the conductor can also be expressed in terms of its mass \(m\) and specific heat capacity \(s\):
\[ m s \frac{dT}{dt} = P - P_{\text{dissipated}} \]
Substituting for \(P\) and \(P_{\text{dissipated}}\):
\[ m s \frac{dT}{dt} = \frac{V^2}{R} - k (T - T_0) \]

6. **Rearranging for Separation of Variables**:
Rearranging gives:
\[ \frac{dT}{\frac{V^2}{R} - k(T - T_0)} = \frac{dt}{m s} \]

7. **Integrating**:
This is a separable differential equation. We can integrate both sides to find \(T(t)\). The left side involves a logarithmic function:
\[ \int \frac{dT}{\frac{V^2}{R} - k(T - T_0)} = \frac{1}{m s} \int dt \]

8. **Solving the Integral**:
The integration results in:
\[ -\frac{1}{k} \ln\left| \frac{V^2/R - k(T - T_0)}{V^2/R} \right| = \frac{t}{m s} + C \]
where \(C\) is the constant of integration determined by initial conditions.

9. **Finding the Constant**:
At \(t = 0\), \(T = T_0\):
\[ C = -\frac{1}{k} \ln\left| 1 \right| = 0 \]

10. **Final Expression**:
Thus, we can express \(T(t)\) as:
\[ T(t) = T_0 + \frac{V^2}{kR} \left(1 - e^{-kt/(ms)}\right) \]

### Part (ii): Find the temperature of the conductor after a long time.

1. **Long Time Condition**:
As \(t \to \infty\), the exponential term \(e^{-kt/(ms)}\) approaches zero.

2. **Final Temperature**:
Therefore, the temperature \(T\) approaches:
\[ T_{\text{final}} = T_0 + \frac{V^2}{kR} \]

### Summary of Results:
- **Part (i)**: The time dependence of the temperature \(T(t)\) is:
\[ T(t) = T_0 + \frac{V^2}{kR} \left(1 - e^{-kt/(ms)}\right) \]
- **Part (ii)**: The final temperature of the conductor after a long time is:
\[ T_{\text{final}} = T_0 + \frac{V^2}{kR} \]