A cylindrical block of length 0.4 m and area of cross-section `0.04 m^2` is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross - section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300K. if the thermal conductivity of the material of the cylinder is `10 "watt"// m-K ` and the specific heat of the material of the disc is `600J//kg-K`, how long will it take for the temperature of the disc to increase to 350 K? Assume for purpose of calculation the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Explanation

To solve the problem, we need to determine how long it will take for the temperature of the metal disc to increase from 300 K to 350 K when it is in contact with a cylindrical block maintained at 400 K. We will use the principles of heat conduction and specific heat capacity.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of the cylinder, \( L = 0.4 \, \text{m} \)
- Cross-sectional area, \( A = 0.04 \, \text{m}^2 \)
- Mass of the disc, \( m = 0.4 \, \text{kg} \)
- Initial temperature of the disc, \( T_i = 300 \, \text{K} \)
- Final temperature of the disc, \( T_f = 350 \, \text{K} \)
- Constant temperature of the cylinder, \( T_c = 400 \, \text{K} \)
- Thermal conductivity of the cylinder, \( k = 10 \, \text{W/m-K} \)
- Specific heat capacity of the disc, \( c = 600 \, \text{J/kg-K} \)

2. **Calculate Heat Transfer Rate (dQ/dt):**
The heat transfer rate through conduction can be expressed as:
\[ \frac{dQ}{dt} = \frac{k \cdot A}{L} (T_c - T) \]
Substituting the values:
\[ \frac{dQ}{dt} = \frac{10 \cdot 0.04}{0.4} (400 - T) = 1 (400 - T) \]
Thus, we have:
\[ \frac{dQ}{dt} = 400 - T \quad \text{(Equation 1)} \]

3. **Relate Heat Transfer to Temperature Change of the Disc:**
The heat gained by the disc can be expressed as:
\[ \frac{dQ}{dt} = m \cdot c \cdot \frac{dT}{dt} \]
Substituting the values:
\[ \frac{dQ}{dt} = 0.4 \cdot 600 \cdot \frac{dT}{dt} = 240 \frac{dT}{dt} \quad \text{(Equation 2)} \]

4. **Set Equations Equal:**
From Equations 1 and 2, we have:
\[ 400 - T = 240 \frac{dT}{dt} \]

5. **Rearrange and Separate Variables:**
Rearranging gives:
\[ \frac{dT}{400 - T} = \frac{dt}{240} \]

6. **Integrate Both Sides:**
Integrating both sides, we have:
\[ \int_{300}^{350} \frac{dT}{400 - T} = \int_{0}^{T} \frac{dt}{240} \]
The left side integrates to:
\[ -\ln(400 - T) \Big|_{300}^{350} = -\ln(400 - 350) + \ln(400 - 300) \]
This simplifies to:
\[ -\ln(50) + \ln(100) = \ln\left(\frac{100}{50}\right) = \ln(2) \]

The right side integrates to:
\[ \frac{T}{240} \]

7. **Equate and Solve for Time (T):**
Thus, we have:
\[ \ln(2) = \frac{T}{240} \]
Solving for \( T \):
\[ T = 240 \ln(2) \]

8. **Calculate \( T \):**
Using \( \ln(2) \approx 0.693 \):
\[ T \approx 240 \times 0.693 \approx 166.32 \, \text{seconds} \]

### Final Answer:
It will take approximately **166.32 seconds** for the temperature of the disc to increase from 300 K to 350 K.