Thermal Conductivity of Rubber

THERMAL CONDUCTIVITY OF RUBBER

Principle

It is based on the principle of radial flow of heat through a cylindrical shell.

Procedure

A big empty calorimeter with stirrer is weighed (W₁). It is then filled with two-thirds of water and again weighed (W₂). A known length (l) of a rubber tube is immersed in water in the calorimeter (fig. 1.10).

The calorimeter is stirred well and the initial temperature θ₁°C is noted.

Now one end of the rubber tube is connected to a steam generator and steam is passed through it.

The steam is passed continuously till there is a rise of 10°C in temperature. The time taken (t second) for this rise in final temperature of the water  (θ₂°C) in the calorimeter is also noted.

Observation

Mass of the empty calorimeter with stirrer = W₁ kg

Mass of the calorimeter + water = W₂ kg

Mass of the water = (W₂- W₁) kg

Initial temperature of the water  = θ₁°C 

Final temperature of the water = θ₂°C

Rise in the temperature of water = (θ_{2 -} θ₁)

Time for which steam is passed = t second

Length of the rubber tube immersed in water = l

Inner radius of the rubber tube = r₁

Outer radius of the rubber tube = r₂

Specific heat capacity of calorimeter = S₁

Specific heat capacity of water = S₂

Heat gained by the calorimeter = Mass ×  specific heat capacity ×  change in temperature

= W₁ × X S₁× (θ₂-θ₁) ………………....(1)

Heat gained by the water =  (W₂- W₁) × S₂ × (θ₂-θ₁) ……………....(2)

Total heat gained by the calorimeter and water in t second

The expression for the thermal conductivity (K) in the case of cylindrical shell method is given by

substituting eqn (3) in eqn (4), we have

where θ_s temperature of steam

θ₁ + θ₂/ 2 -Average temperature inside of rubber tube.

Problem 1.5

2demiolo lo viosqes ised oftiosq A metal pipe having an external diameter 20 cm carries steam at 100°C. This is covered by a layer 2.0 cm thick of insulating material with co-efficient of thermal conductivity 0.20 Wm^-1 K^-1. If the outer surface is 30°C, calculate the heat lost by the pipe of 2m length per hour. Neglect the temperature drop across the pipe. (A.U. Nov. 2019)

Given data:

Solution:

Formula:

Quantity of heat flowing out across the pipe of length 'l' m in time 't' seconds

∴ The quantity of heat lost by the pipe per hour Q = Q = 3.472 × 10⁶ Joules