Two beakers, identical in all respects, but made of different materials, are filled with equal amounts of ice at `0^(@) C`. Toe ice in one beaker melts in one minute whereas that in the other beaker melts in five minutes. Find the ratio of the thermal conductivities of the material of the two beakers.

Explanation

To solve the problem of finding the ratio of the thermal conductivities of the materials of the two beakers, we can follow these steps:

### Step 1: Understand the problem
We have two beakers made of different materials, both containing equal amounts of ice at 0°C. One beaker melts the ice in 1 minute, while the other melts it in 5 minutes. We need to find the ratio of the thermal conductivities of the materials of the two beakers.

### Step 2: Use the formula for heat transfer
The heat transfer \( Q \) through a material can be expressed using the formula:
\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]
where:
- \( K \) is the thermal conductivity of the material,
- \( A \) is the area through which heat is being transferred,
- \( \Delta T \) is the temperature difference,
- \( L \) is the thickness of the material.

### Step 3: Set up the equations for both beakers
Let’s denote:
- For Beaker 1 (melts in 1 minute): \( K_1, Q_1, T_1 = 1 \text{ min} \)
- For Beaker 2 (melts in 5 minutes): \( K_2, Q_2, T_2 = 5 \text{ min} \)

Since both beakers are identical in all respects except for the material, we can assume \( A \) and \( L \) are the same for both beakers. The temperature difference \( \Delta T \) is also the same (since both are at 0°C).

### Step 4: Relate the heat transfer to time
The amount of heat transferred \( Q \) is also related to the time taken for the ice to melt:
\[ \frac{Q_1}{T_1} = \frac{Q_2}{T_2} \]
This means that the heat transferred per unit time is proportional to the thermal conductivity.

### Step 5: Establish the proportional relationship
From the heat transfer equation, we can write:
\[ \frac{K_1 \cdot A \cdot \Delta T}{L \cdot T_1} = \frac{K_2 \cdot A \cdot \Delta T}{L \cdot T_2} \]
Since \( A, \Delta T, \) and \( L \) are constant, we can simplify this to:
\[ \frac{K_1}{T_1} = \frac{K_2}{T_2} \]

### Step 6: Solve for the ratio of thermal conductivities
Rearranging gives us:
\[ \frac{K_1}{K_2} = \frac{T_1}{T_2} \]
Substituting the values \( T_1 = 1 \) min and \( T_2 = 5 \) min:
\[ \frac{K_1}{K_2} = \frac{1}{5} \]
Thus, the ratio of the thermal conductivities is:
\[ K_1 : K_2 = 1 : 5 \]

### Final Answer
The ratio of the thermal conductivities of the materials of the two beakers is \( 1 : 5 \).

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