A uniform cylinder of length `L` and thermal conductivity `k` is placed on a metal plate of the same area `S` of mass `m` and infinite conductivity. The specific heat of the plate is `c`. The top of the cylinder is maintained at `T_(0)`. Find the time required for the temperature of the plate to rise from `T_(1)` to `T_(2)(T_(1) lt T_(2) lt T_(0))`.

Explanation

To solve the problem step by step, we will analyze the heat transfer from the cylinder to the metal plate and derive the time required for the temperature of the plate to rise from \( T_1 \) to \( T_2 \).

### Step 1: Understanding the System
We have a uniform cylinder of length \( L \) and thermal conductivity \( k \) placed on a metal plate with mass \( m \), area \( S \), and specific heat \( c \). The top of the cylinder is maintained at temperature \( T_0 \), and we want to find the time required for the plate's temperature to rise from \( T_1 \) to \( T_2 \).

### Step 2: Heat Transfer Equation
The rate of heat transfer \( \frac{dq}{dt} \) through the cylinder can be expressed using Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{kA(T_0 - T)}{L} \]
where \( T \) is the temperature of the plate at any time \( t \).

### Step 3: Relating Heat Transfer to Temperature Change
The heat absorbed by the metal plate can also be expressed in terms of its mass and specific heat:
\[ \frac{dq}{dt} = mc \frac{dT}{dt} \]
where \( T \) is the temperature of the plate.

### Step 4: Setting Up the Equation
Equating the two expressions for \( \frac{dq}{dt} \):
\[ mc \frac{dT}{dt} = \frac{kS(T_0 - T)}{L} \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ \frac{dT}{T_0 - T} = \frac{kS}{mcL} dt \]

### Step 6: Integrating Both Sides
Now we will integrate both sides. The left side will be integrated from \( T_1 \) to \( T_2 \), and the right side from \( 0 \) to \( t \):
\[ \int_{T_1}^{T_2} \frac{dT}{T_0 - T} = \frac{kS}{mcL} \int_0^t dt \]

### Step 7: Performing the Integration
The left side integrates to:
\[ -\ln(T_0 - T) \bigg|_{T_1}^{T_2} = -\ln(T_0 - T_2) + \ln(T_0 - T_1) = \ln\left(\frac{T_0 - T_1}{T_0 - T_2}\right) \]
The right side integrates to:
\[ \frac{kS}{mcL} t \]

### Step 8: Setting the Equations Equal
Setting both integrals equal gives us:
\[ \ln\left(\frac{T_0 - T_1}{T_0 - T_2}\right) = \frac{kS}{mcL} t \]

### Step 9: Solving for Time \( t \)
Now, we can solve for \( t \):
\[ t = \frac{mcL}{kS} \ln\left(\frac{T_0 - T_1}{T_0 - T_2}\right) \]

### Final Result
Thus, the time required for the temperature of the plate to rise from \( T_1 \) to \( T_2 \) is:
\[ t = \frac{mcL}{kS} \ln\left(\frac{T_0 - T_1}{T_0 - T_2}\right) \]