UNIT 13 : HEAT

1. UNIT 13 : HEAT 13.1 Thermal Conductivity 13.2 Thermal Expansion

2. 13.1 Thermal ConductivityAt the end of this topic, students should be able to: • Define heat as energy transfer due to temperature difference. • Explain the physical meaning of thermal conductivity. • Use rate of heat transfer, • Use temperature-distance graphs to explain heat conduction through insulated and non-insulated rods, and combination of rods in series.

3. Heat • is defined as the energy that is transferred from a body at • a higher temperature to one at a lower temperature , by • conduction, convection or radiation. • Heat always transferred from a hot region (higher • temperature) to a cool region (lower temperature) until • thermal equilibrium is achieved. • Heat is transferred by three mechanisms, • 1) Conduction • 2) Convection • 3) Radiation • Thermal Conduction is defined as the process whereby • heat is transferred through a substance from a region • of high temperature to a region of lower temperature.

4. B A The mechanism of heat conduction through solid material (for extra knowledge only) • Suppose a rod is heated at one end (A). • Before the rod being heated all the molecules vibrate • about their equilibrium position. • As the rod is heated the molecules at the hot end (A) • vibrate with increasing amplitude, thus the kinetic energy • increases. • While vibrating the hot molecules collide with the • neighbouring colder molecules result in transfer of kinetic • energy to the colder molecules. • This transfer of energy will continue until the cold end (B) • of the rod become hot.

5. Thermal conductivity, k • Consider a uniform cylinder conductor of length l with • temperature T1 at one end and T2at the other end as • shown in figure above. • The heat flows to the right because T1 is greater than T2.

6. The rate of heat flow , • through the conductor is given by:

7. The rate of of heat flow through an object depends on : • Thermal conductivity. • Cross-sectional area through which the heat flow. • Thickness of the material. • Temperature difference between the two sides of the material.

8. The negative sign because the temperature T, • become less as the distance, x increases. • The rate of heat flow is a scalar quantity • and its unit is J s-1 or Watt (W). Thermal conductivity , k is defined as the rate of heat flows perpendicularly through unit cross sectional area of a solid , per unit temperature gradient along the direction of heat flow. Thermal conductivity is a property of conducting material. ( the ability of the material to conduct heat) where good conductors will have higher values of kcompared to poor conductors.

9. Materials with large k are called conductors; those with small k are called insulators.

10. Heat conduction through insulated rod • Consider heat conduction through an insulated rod which • has cross sectional area A and length xas shown above. • If the rod is completely covered with a good insulator, no • heat loss from the sides of the rod. • By assuming no heat is lost to the surroundings, therefore • heat can only flow through the cross sectional area from • higher temperature region, T1 to lower temperature region, • T2.

11. insulator insulator • The red lines (arrows) represent the direction of heat flow. • When the rod is in steady state (the temperature falls at • a constant rate) thus the rate of heat flows is constant along • the rod. • This causes the temperature gradient will be constant along • the rod as shown in figures above.

12. Heat conduction through non-insulated rod • The metal is not covered with an insulator, thus heat is • lost to the surroundings from the sides of the rod. • The lines of heat flow are divergent and the temperature • fall faster near the hotter end than that near the colder end.

13. X dT dx Y • Less heat is transferred to Y. • This causes the temperature gradient gradually • decreases along the rod and result a curve graph where • the temperature gradient at X higher than that at Y as • shown in figure below. • From • And from the graph where A and k are the same along the rod. • Thus Temperature gradient , at any point on the rod is given by the slope of the tangent at that point.

14. insulator and Material C Material D insulator Combination 2 metals in series xc xD

15. When steady state is achieved , the rate of heat • flow through both materials is same (constant). • From the equation of thermal conductivity, we get But

16. Example 13.1 A metal cube have a side of 8 cm and thermal conductivity of 250 W m-1 K-1. If two opposite surfaces of the cube have the temperature of 90 C and 10 C, respectively. Calculate a) the temperature gradient in the metal cube. b) the quantity of heat flow through the cube in 10 minutes. (Assume the heat flow is steady and no energy is lost to the surroundings)

17. Solution 13.1 A = l2 = (8x102)2= 64 x104 m2, k = 250 W m-1 K-1, T1= 90C, T2 = 10C b)Givent= 10 x 60 = 600 s

18. Example 13.2 A 5 mm thick copper plate is sealed to a 10 mm thick aluminium plate and both have the same cross sectional area of 1 m2.The outside face of the copper plate is at 100 C, while the outside face of the aluminium plate is at 80 C. a) Find the temperature at the copper-aluminium interface. b) Calculate the rate of heat flow through the cross sectional area if heat flow is steady and no energy is lost to the surroundings. (Use kCu = 400 W m-1C-1 and kAl = 200 W m-1C-1)

19. Solution 13.2 xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C • The rate of heat flow • through the copper and aluminium plate is same, therefore

20. Solution 13.2 xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C

21. Solution 13.2 xCu= 5x10-3m, xAl= 10x10-3m, A= 1 m2, Tcu = 100C, TAl = 80C b) By applying the equation for rate of heat flow through the copper plate, hence

22. Exercise 1. A metal plate 5.0 cm thick has a cross sectional area of 300 cm2. One of its face is maintained at 100C by placing it in contact with steam and another face is maintained at 30C by placing it in contact with water flow. Determine the thermal conductivity of the metal plate if the rate of heat flow through the plate is 9 kW. (Assume the heat flow is steady and no energy is lost to the surroundings).( 214 W m-1K-1 ) 2. A rod 1.300 m long consists of a 0.800 m length of aluminium joined end to end to a 0.500 m length of brass. The free end of the aluminium section is maintained at 150.0C and the free end of the brass piece is maintained at 20.0C. No heat is lost through the sides of the rod. At steady state, find the temperature of the point where the two metal are joined.(Use k of aluminium = 205 W m-1C-1 and k of brass = 109 W m-1C-1) (90.2C)

23. 13.2 Thermal expansionAt the end of this topic, students should be able to: • Define and use the coefficient of linear, area and volume thermal expansion. • Deduce the relationship between the coefficients of expansion , .

24. 14.3 Thermal expansion • Thermal expansion is defined as the change in • dimensions of a body accompanying a change in • temperature. • 3 types of thermal expansion : • - Linear expansion • - Area expansion • - Volume expansion • In solid, all types of thermal expansion are occurred. • In liquid and gas, only volume expansion is occurred. • At the same temperature, the gas expands greater than • liquid and solid.

25. Linear expansion • Consider a thin rod of initial length, l0 at temperature,T0 is heated to a new uniform temperature, T and acquires length, l as shown in figure below. • If ΔT is not too large (< 100o C) and

26. Unit of  is C-1 or K-1. • The coefficient of linear expansion,  is defined as • the change in length of a solid per unit length per unit • rise change in temperature. • If the length of the object at a temperature T is l, • For many materials, every linear dimension changes • according to both equations above. Thus, l could be the • length of a rod, the side length of a square plate or the • diameter (radius) of a hole.

27. For example, as a metal washer is heated, all dimensions • including the radius of the hole increase as shown in • figure below.

28. Area expansion • This expansion involving the expansion of a surface area of • an object. • Consider a plate with initial area, A0 at temperature T0 is • heated to a new uniform temperature, T and expands by A, • as shown in figure below. • From this experiment, we get and

29. Unit of  is C-1 or K-1 • The coefficient of area expansion,  is defined as the • change in area of a solid surface per unit area per • unit rise in temperature. • The area of the of the surface of object at a • temperature T can be written as, • For isotropic material (solid) , the area expansion is • uniform in all direction, thus the relationship between  and •  is given by

30. Derivation • Consider a square plate with side length, l0 is heated and expands uniformly as shown in figure below. because where and compare with

31. Volume expansion • Consider a metal cube with side length, l0 is heated and expands uniformly. From the experiment, we get and Unit of  is C-1 or K-1. The coefficient of volume expansion,  is defined as the change in volume of a solid per unit volume per unit rise in temperature.

32. Δl lo l • The volume of an object at a temperature T can be • written as, • For isotropic material (solid), the volume expansion is • uniform in all direction, thus the relationship between  • and  is given by Derivation • Consider a metal cube with side length, l0 is heated and expands uniformly.

33. because where and compare with

34. Example 13.3 The length of metal rod is 30.000 cm at 20C and 30.019 cm at 45C, respectively. Calculate the coefficient of linear expansion for the rod. Solution l0= 30.000 cm, T0= 20C, l= 30.019 cm, T = 45C

35. Example 13.4 A steel ball is 1.900 cm in diameter at 20.0C. Given that the coefficient of linear expansion for steel is 1.2 x 105C-1, calculate the diameter of the steel ball at a) 57.0C b) 66.0C Solution d0= 1.900 cm, T0= 20.0C,  = 1.2x105C-1 a) b)

36. Example 13.5 A cylinder of radius 18.0 cm is to be inserted into a brass ring of radius 17.9 cm at 20.0C. Find the temperature of the brass ring so that the cylinder could be inserted. (Given the coefficient of area expansion for brass is 4.0 x 105C-1) Solution rc= 18.0 cm, r0b= 17.9 cm, T0= 20.0C,  = 4.0x105C-1 When the cylinder pass through the brassring, thus

37. Example 13.6 Determine the change in volume of block of cast iron 5.0 cm x 10 cm x 6.0 cm, when the temperature changes from 15 oC to 47 oC. ( = 0.000010 oC -1 ) Solution

38. Exercise • A rod 3.0 m long is found to have expanded 0.091 cm in • length after a temperature rise of 60 o C. What is the coefficient of linear expansion for the material of the rod ? 5.1 x 106o C1 2. The length of a copper rod is 2.001 m and the length of a wolfram rod is 2.003 m at the same temperature. Calculate the change in temperature so that the two rods have the same length where the final temperature for both rods is equal. (Given the coefficient of linear expansion for copper is 1.7 x 105C1 and the coefficient of linear expansion for wolfram is 0.43 x 105C1) 78.72C

39. Thermal Expansion of Liquid in A Container • When a liquid in a solid container is heated, both liquid and • the solid container expand in volume. • Liquid expands more than the solid container. • The change in volume of a liquid is given by • The coefficient of volume expansion of a liquid is defined in • the same way as the coefficient of volume of a solid i.e :

40. Example 13.7 A glass flask with a volume of 200 cm3 is filled to the brim with mercury at 20 oC. How much mercury overflows when the temperature of the system is raised to 100 oC ? Solution