Two vessels 1 and 2, made out of different materials are identical in all the geometrical aspects. In both the vessel the same quantity of ice gets melted in 20 min and 30 min respectively. The ratio of the thermal conductivity of the second one to that of the first is

Options

A. 1.5
B. 1
C. `2/3`
D. `4`

Answer

To solve the problem, we will analyze the heat transfer through the two vessels and use the relationship between thermal conductivity and the time taken for the ice to melt.

### Step-by-step Solution:

1. **Understanding Heat Transfer**:
The rate of heat transfer through a material is given by the formula:
\[ \frac{dQ}{dt} = \frac{k \cdot A \cdot \Delta T}{L} \]
where:
- \(dQ\) is the heat transferred,
- \(dt\) is the time,
- \(k\) is the thermal conductivity,
- \(A\) is the area,
- \(\Delta T\) is the temperature difference,
- \(L\) is the thickness of the material.

2. **Identifying Constants**:
In this problem, both vessels are identical in geometry, which means:
- The area \(A\) is the same for both vessels.
- The thickness \(L\) is the same for both vessels.
- The temperature difference \(\Delta T\) is also the same since both vessels are melting the same quantity of ice at the same surrounding temperature.

3. **Heat Required to Melt Ice**:
The heat required to melt a mass \(m\) of ice is given by:
\[ Q = m \cdot L_f \]
where \(L_f\) is the latent heat of fusion. Since the same quantity of ice is melted in both vessels, the heat \(Q\) is the same for both.

4. **Relating Time and Thermal Conductivity**:
Since the heat \(Q\) is constant, we can relate the time taken for the ice to melt in each vessel to the thermal conductivity:
\[ \frac{Q}{t_1} = \frac{k_1 \cdot A \cdot \Delta T}{L} \quad \text{(for vessel 1)} \]
\[ \frac{Q}{t_2} = \frac{k_2 \cdot A \cdot \Delta T}{L} \quad \text{(for vessel 2)} \]
Rearranging gives:
\[ k_1 = \frac{Q \cdot L}{A \cdot \Delta T \cdot t_1} \]
\[ k_2 = \frac{Q \cdot L}{A \cdot \Delta T \cdot t_2} \]

5. **Finding the Ratio of Thermal Conductivities**:
Taking the ratio of \(k_2\) to \(k_1\):
\[ \frac{k_2}{k_1} = \frac{t_1}{t_2} \]
Given \(t_1 = 20 \text{ min}\) and \(t_2 = 30 \text{ min}\):
\[ \frac{k_2}{k_1} = \frac{20}{30} = \frac{2}{3} \]

6. **Final Answer**:
The ratio of the thermal conductivity of the second vessel to that of the first is:
\[ \frac{k_2}{k_1} = \frac{2}{3} \]