**Object.** To determine the thermal conductivity of a non-conducting material by the Lee’s disc method.

**Apparatus Required.** Lee’s apparatus, steam boiler, two thermometers, vernier callipers, heater, stop-watch, screw gauge, physical balance and weight box.

**Description of the apparatus.** Lee’s apparatus is shown in Fig. 1. It consists of a metallic ring R attached to a heavy vertical stand. A circular solid disc C of brass (or copper) is suspended from the ring R. A steam chamber S is kept on the disc C. The steam chamber is provided with another solid disc B at its base. The experimental non-conducting disc G is kept in between the discs B and C. The discs B and C have openings to insert thermometers T_{1}, and T_{2} in them.

**Theory.** For the experiment, steam is passed in the steam chamber S. In the steady state, the amount of heat passing per second from the disc B to the disc C through the non-conducting disc G, is lost per second by radiation from the lower surface and curved surface of the disc C. Let A=п*r*^{2} if *r* is the radius of disc) be the area of plane surface of experimental disc and *x* be its thickness. If in steady state, the temperature recorded by the thermometers T_{1}, and T_{2} be θ_{1} and θ_{2 }respectively, then heat conducted per second through the non-conducting disc is

\[

Q = \frac{KA(\theta_1 – \theta_2)}{x} … (1)

\]

where K is the coefficient of thermal conductivity of the material of nonconducting disc.

Now if the rate of fall of temperature of disc C at temperature θ_{2 }is *(dθ/dt) _{θ=θ2}* and the mass of disc C is

*m*, specific heat of its material is

*s*, then the heat lost per second by radiation from the disc C is

\[

Q’ = ms \left( \left. \frac{d\theta}{dt} \right|_{\theta_0} \right)_{\theta_2} … (2)

\]

In the steady state, if the heat radiated by the disc G is neglected then **Q = Q’**

then,

\begin{equation}

\frac{KA(\theta_1 – \theta_2)}{x} = ms \left( \left. \frac{d\theta}{dt} \right|_{\theta_0=\theta_2} \right)

\end{equation}

\begin{equation}

K = \frac{ms \left( \frac{d\theta}{dt} \right)_{\theta_0=\theta_2}}{A(\theta_1 – \theta_2)} = \frac{ms \left( \frac{d\theta}{dt} \right)_{\theta_0=\theta_2}}{\pi r^2(\theta_1 – \theta_2)}. … (3)

\end{equation}

The value of K can be calculated from the above expression.

**Formula used.** Coefficient of thermal conductivity

\[

K = \frac{msx \left( \left. \frac{d\theta}{dt} \right|_{\theta_0=\theta_2} \right)}{\pi r^2 (\theta_1 – \theta_2)}

… (4)

\]

where,

*m* = mass of disc C,

*s* = specific heat of the material of disc C,

*x *= thickness of experimental disc,

*r* = radius of experimental disc,

*(dθ/dt) _{θ=θ2}* = rate of fall of temperature of disc at temperature θ, θ

_{1}– θ

_{2}= difference in temperatures of the surfaces of disc in the steady state.

[Remember that if *m* is in gm, *s* in cal/gm °C, *x* and *r* in cm, then the value of K will be in cal/sec cm °C].

**Procedure.** **(Not to be written in copy)**

1. First fill water in the steam boiler and place it on the burner to form steam. Then measure the mass *m* of the lower disc C with the help of physical balance.

2. Note least count and zero error of vernier callipers. Measure the diameter of the experimental non-conducting disc G at various places and at each place in two mutually perpendicular directions, using a vernier callipers. Then take its mean and divide by 2 to find its mean radius *r*.

3. Note least count and zero error of screw gauge. Measure the thickness of the experimental non-conducting disc G at several places with the help of screw gauge.

4. Now place the experimental disc G in between the discs B and C. Insert the thermometers T_{1} and T_{2} in the openings provided in discs B and C.

5. Allow steam to pass in the steam chamber from the boiler. Wait for 4-5 minutes and then note the readings of thermometers T_{1} and T_{2} after each 2 minutes. When the readings θ_{1} and θ_{2} become constant, note these readings.

6. Stop passing steam in the steam chamber and remove the disc C from the apparatus. Now heat the disc C by keeping it on a heater till its temperature rises to 10°C above θ_{2}.Suspend the disc C alone from the ring R and allow it to cool. Now, record its cooling by noting the time (in seconds) for the fall in temperature by 1°C, till its temperature falls to 10°C below θ_{2}. Note the specific heat of material of disc C form the standard table.

7. Then plot a graph X-axis and temperature θ on Y-axis as shown in Fig. 2. Draw tangent to the curve at point P corresponding to the temperature θ_{2} and find its slope which gives the value of *(dθ/dt) _{θ=θ2}*.

**Observations.**

1. Specific heat of material of disc C, *s* = ………….. cal/gm °C (From standard tables)

2. Mass of disc C, *m =*……….. gm

3. **For the radius of non-conducting disc G**

Least count of vernier callipers = ……. cm

Zero error = ± …… cm

S.No. | Main Scale Reading (cm) | Vernier Scale Reading (cm) | Total reading = Observed diameter (cm) |

1. Along one direction Along ┴ direction | |||

2 Along one direction Along ┴ direction | |||

3. Along one direction Along ┴ direction | |||

Mean observed diameter = ……..cm

Mean diameter *d* of disc G = Observed diameter – Zero error (with sign)

=………… cm

Radius of disc G, *r* = = ……… cm

4. **For the thickness of disc G**

Least count of screw gauge = …… cm

Zero error = ± …… cm

S. No. | Main Scale Reading (cm) | Circular Scale Reading (cm) | Total reading (cm) |

1. | |||

2. | |||

3. |

Mean observed thickness = ……..cm

Actual thickness of disc G, x = observed thickness – zero error (with sign)

=…….cm

5. Least count of thermometer = ……°C

6. Steady temperature of thermometer T_{1} (i.e., steady temperature of disc B) θ_{1}=……….⁰C

7. Steady temperature of thermometer T_{2} (i.e., steady temperature of disc C) θ_{2}=………⁰C

8. **For the cooling curve of disc C**

Least count of stopwatch = …………. sec

S.No. | Time t (sec) | Temperature of disc C (⁰C) | S.No. | Time t (sec) | Temperature of disc C (⁰C) |

1. | 11. | ||||

2. | 12. | ||||

3. | 13. | ||||

4. | 14. | ||||

5. | 15. | ||||

6. | 16. | ||||

7. | 17. | ||||

8. | 18. | ||||

9. | 19. | ||||

10. | 20. |

9. From the above observation table, a graph is plotted for temperature θ on Y-axis and time *t* on X-axis. From this graph (Fig. 2), the slope of the tangent drawn on curve at θ = θ_{2}, is

*(dθ/dt) _{θ=θ2} *

**=**

**=**

**…..°C/sec.**

**Calculations.** Substituting the values of all the quantities on the right-hand side of the expression

the value of K can be calculated.

K = ………………… cal/sec cm °C.

**Result.** The coefficient of thermal conductivity of the given non-conducting material (………………….) = ………….. cal cm^{-1} sec^{-1} °C^{-1}

Standard value = ………….. cal cm^{-1} sec^{-1} °C^{-1} (From standard tables)

**Precautions.**

- The readings of thermometers should be noted after attaining the steady state.
- The thermometers must be sensitive.
- The experimental non-conducting disc must be thin so that heat loss by radiation from its curved surface may be negligible.
- The thickness and diameter of the experimental non-conducting disc must be measured before placing it in the apparatus (i.e., before heating it).
- Before the experiment, it must be checked that there is sufficient water in the boiler for generation of steam.

__Viva-Voce__

**Q. What are the various modes of transfer of heat?**

Ans. (i) conduction, (ii) convection, and (iii) radiation.

**Q. How is heat transferred particularly in solids?**

Ans. By conduction.

**Q. What do you understand by the terms ‘thermal conductivity’ and ‘coefficient of thermal conductivity’?**

Ans. Thermal conductivity is the property of a substance due to which heat is transferred through that substance by conduction. The coefficient of thermal conductivity of a material is the amount of heat passing per second through the rod of that material of length 1 cm and area of cross-section 1 cm^{2} when the difference in temperatures at its ends is 1°C.

**Q. What is the unit of coefficient of thermal conductivity?**

Ans. cal cm^{-1} sec^{-1}°C^{-1}or joule metre^{-l} second^{-1} °C^{-1}.

**Q. Which substance is best conductor of heat?**

Ans. Silver.

**Q. What are the values of K for an ideal conductor and an ideal insulator?**

Ans. K = infinite for an ideal conductor and K = zero for an ideal insulator.

**Q. Which of the following is the best conductor of heat? Copper, Silver, Aluminium, Brass.**

Ans. Silver.

**Q. Which of the following is the best insulator of heat? Wood, Glass, Mercury, Air.**

Ans. Air.

**Q. Name a substance which is good conductor of heat but an insulator of electricity?**

Ans. Mica.

**Q. On what factors does the rate of rise in temperature of disc in the variable state depend?**

Ans. On (i) thermal conductivity, (ii) specific heat, and (iii) density of the material of disc.

**Q. Why do the readings of the two thermometers not change with time in the steady state?**

Ans. Because in the steady state, the rate of flow of heat from the B to the disc C through the non-conducting disc is equal to the rate at which heat is radiated from the lower surface and curved surface of the disc C.

**Q. Do the two thermometers have the same readings, in the steady state?**

Ans. No. The two thermometers do not have the same readings, but there is a temperature gradient between the plane surfaces of the experimental disc.

**Q. What do you mean by the term temperature gradient?**

Ans. The rate of fall of temperature with distance in direction of flow of heat is called the temperature gradient.

**Q. On what factors does the amount of heat conducted through the disc in the steady state depends?**

Ans. In the steady state, the amount of heat Q conducted through the disc is (i) directly proportional to the area of cross-section A of the disc, (ii) inversely proportional to the thickness *x* of disc, (iii) directly proportional to the temperature difference θ_{1 }– θ_{2 }across the disc, and (iv) directly proportional to time *t*.

**Q. You are given two circular discs of same thickness and same radius with the same temperature difference across them. But one disc is of glass and the**

**other disc is of rubber. Will the rate of heat conducted by them be equal?**

Ans. No. The rate of conduction of heat through the glass disc will be more than that through the rubber disc.

**Q. Why do you take the experimental substance in the form of circular disc in this experiment?**

Ans. So that the thickness of disc is small, but its area of cross-section is large (otherwise the rate of conduction of heat through the non-conducting material will be very low).

**Q. Can you determine the heat conductivity of a good conductor this method?**

Ans. No. The reason is that if we take the disc of a good conductor, the rate of conduction of heat through it will be very high and so the temperatures across it will be nearly equal.

**Q. Can you determine the conductivity of a liquid by this method?**

Ans. Yes. For this, the experimental liquid is taken in a hollow thin-walled disc of a non-conducting material and this disc is then placed between the discs B and C.

**Q. Why do you note the temperature recorded by the thermometers when the steady state is attained?**

Ans. Because while obtaining the expression used in this experiment, it has been assumed that the experimental disc is in the steady state.

**Q. How do you check that the steady state has been achieved?**

Ans. Steady state is achieved when the temperatures recorded by the two thermometers do not change with time.

**Q. How is the coefficient of thermal conductivity of a substance affected by the change in temperature?**

Ans. Generally the coefficient of thermal conductivity of a substance decreases with the increase in its temperature.