A rod of negligible heat capacity has length 20cm, area of cross section `1.0cm^(2)` and thermal conductivity `200Wm^(-1)C^(-1)` . The temperature of one end is maintained at `0^(@)C` and that of the other end is slowly and linearly varied from `0^(@)C` to `60^(@)C` in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

Explanation

To solve the problem step by step, we will use the formula for heat transfer through conduction and the properties of arithmetic progression.

### Step 1: Identify Given Data
We have the following data:
- Length of the rod, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Area of cross-section, \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)
- Thermal conductivity, \( k = 200 \, \text{W/m°C} \)
- Temperature at one end, \( \theta_1 = 0°C \)
- Temperature at the other end varies from \( \theta_2 = 0°C \) to \( 60°C \) over \( t = 10 \, \text{minutes} = 600 \, \text{seconds} \)

### Step 2: Calculate the Temperature Gradient
The temperature gradient \( \frac{d\theta}{dt} \) can be calculated as:
\[ \frac{d\theta}{dt} = \frac{\Delta \theta}{\Delta t} = \frac{60°C - 0°C}{600 \, \text{s}} = 0.1 \, \text{°C/s} \]

### Step 3: Write the Heat Transfer Equation
The heat transfer \( \frac{dq}{dt} \) through the rod can be expressed using Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{kA(\theta_2 - \theta_1)}{L} \]
Since \( \theta_1 = 0°C \) and \( \theta_2 \) changes from \( 0°C \) to \( 60°C \), we can express \( dq \) as:
\[ dq = \frac{kA}{L} \cdot (\theta_2) \cdot dt \]

### Step 4: Calculate Total Heat Transmitted
To find the total heat transmitted over the 10 minutes, we need to integrate the heat transfer over time. Since \( \theta_2 \) varies linearly, we can sum the contributions from \( 0°C \) to \( 60°C \).

The total heat \( Q \) can be calculated as:
\[ Q = \int_0^{600} \frac{kA}{L} \cdot (0.1t) \, dt \]
This integral represents the sum of heat transferred at each second from \( 0 \) to \( 600 \) seconds.

### Step 5: Solve the Integral
Calculating the integral:
\[ Q = \frac{kA}{L} \cdot 0.1 \int_0^{600} t \, dt \]
The integral \( \int_0^{600} t \, dt = \frac{t^2}{2} \bigg|_0^{600} = \frac{600^2}{2} = 180000 \).

Now substituting back:
\[ Q = \frac{200 \times 1 \times 10^{-4}}{0.2} \cdot 0.1 \cdot 180000 \]
\[ Q = \frac{200 \times 10^{-4}}{0.2} \cdot 0.1 \cdot 180000 = 1000 \cdot 0.1 \cdot 180000 = 1800 \, \text{J} \]

### Final Answer
The total heat transmitted through the rod in 10 minutes is \( \boxed{1800 \, \text{J}} \).