Question

Two vessels of different material are identical in size and in dimension.They are filled with equal quantity of ice at 0C. If ice in both vessels melts completely in 15 minutes and 10 minutes, compare the thermal conductivity of the metals of both vessels?

Answer 1

Two vessels of different material are identical in size and in dimension. They are filled with equal quantity of ice at 0C. If ice in both vessels melts completely in 15 minutes and 10 minutes, compare the thermal conductivity of the metals of both vessels?

**Solution.**

$$\Delta t _ {1} = 1 5 \min , \Delta t _ {2} = 1 0 \min ;$$ $$\frac {k _ {2}}{k _ {1}} -?$$

The integral form of Fourier's law:

$$\frac {\Delta Q}{\Delta t} = k S \frac {\Delta T}{\Delta x};$$

$\Delta Q$ - the heat that is required to melt of the ice;

$\Delta t$ - the time of melting of the ice;

$k$ - the metal's thermal conductivity;

$S$ - the surface area of the vessel;

$\Delta T$ - the temperature difference between environment and vessel;

$\Delta x$ - the wall thickness of the vessel.

The integral form of Fourier's law for the first vessel:

$$\frac {\Delta Q _ {1}}{\Delta t _ {1}} = k _ {1} S _ {1} \frac {\Delta T _ {1}}{\Delta x _ {1}}.$$

The integral form of Fourier's law for the second vessel:

$$\frac {\Delta Q _ {2}}{\Delta t _ {2}} = k _ {2} S _ {2} \frac {\Delta T _ {2}}{\Delta x _ {2}}.$$

The heat required for melting the ice in the first vessel is the same as in the second vessel then:

$$\Delta Q _ {1} = \Delta Q _ {2} = \Delta Q.$$

The temperature difference between environment and the first vessel is the same as the temperature difference between environment and the second vessel then:

$$\Delta T _ {1} = \Delta T _ {2} = \Delta T.$$

Two vessels are identical in size.

The surface area of the first vessel is the same as the surface area of the second vessel then:

$$S _ {1} = S _ {2} = S.$$

The wall thickness of the first vessel is the same as the wall thickness of the second vessel then:

$$\Delta x _ {1} = \Delta x _ {2} = \Delta x.$$

There are two equations:

First equation:

$$\frac {\Delta Q}{\Delta t _ {1}} = k _ {1} S \frac {\Delta T}{\Delta x}.$$

Second equation:

$$\frac {\Delta Q}{\Delta t _ {2}} = k _ {2} S \frac {\Delta T}{\Delta x}.$$

After dividing the second equation by the first, we get:

$$\frac {\Delta t _ {1}}{\Delta t _ {2}} = \frac {k _ {2}}{k _ {1}}.$$

The ratio of the thermal conductivity of the metals of both vessels is:

$$\frac {k _ {2}}{k _ {1}} = \frac {\Delta t _ {1}}{\Delta t _ {2}}.$$ $$\frac {k _ {2}}{k _ {1}} = \frac {15 \text{min}}{10 \text{min}} = 1.5.$$

Answer: The ratio of the thermal conductivity of the metals of both vessels is $\frac{k_2}{k_1} = 1.5$.