THERMAL PROPERTIES OF MATTER

1. A steel bar (200m) long 20 . If the extremes of temperature to which it might be exposed are -30 to +40 how much will it constract and expand? Given the coefficient of linear expansion of steel is 12 .α ˚

Lo=200m T=20 TF = 40 coefficient= 12x10 K
For the expansion

L = Lo T
=(12x10 K )(200m)(40-20)
=4.8x10 m

For the contraction

L = Lo T
=(12x10 K )(200m)(-30-20)
=12x10 m
Or the decreace in length is 12.0cm



2. A cooper plate (α=1.70 Xˉ⁵Kˉ¹) has dimension of 80cmx15cm at 15˚C. what is the decrease in area when it heated to 35˚C.

α=1.70 Xˉ⁵Kˉ¹ Ao=80cmx15cm T=15˚C TF =35˚C
A= Ao (1+2 α ΔT)
= 0.12(1+2(1.70 Xˉ⁵Kˉ¹)(35˚C-15˚C)
= 0.12m²




3. A circular cooper rng at 20.0˚C has a hole with an area of 9.98cm². what
Minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross sectional area of 10.0cm²? Given α=1.70 Xˉ⁵Kˉ¹

ΔA=2 αAoΔT
10.0cm²-9.98cm²= 2(1.70 Xˉ⁵Kˉ¹)( 9.98cm²)(ΔT)
0.2 = 3.39x10ˉ⁴ΔT

ΔT = (0.2 / 3.39 x10ˉ⁴)
= 58.9 ˚C





4. Two metal rods, one aluminium and one brass, are each comped at one. At 0.0˚C the rods are each 50.0cm long are separated by 0.024cm their unfastened ends. At what temperature will the rods just come into contract.

Lo =50.0cm To = 0.0˚C
αbr = 19x10ˉ⁶Kˉ¹
αAl = 2.28x10ˉ⁶Kˉ¹
αbr+αA l =19x10ˉ⁶Kˉ¹+2.28x10ˉ⁶Kˉ¹
= 0.024cm
ΔT = 0.024cm / (19X10ˉ⁶ + 2.25 X 10ˉ⁶)(50.0cm)

= 0.024cm / (2.075 X 10ˉ⁵)

ΔT = 11.56˚C

TF = To +ΔT
= 0.0˚C + 11.56˚C
= 12˚C





5. A quartz sphere is 8.75cm in diameter. What will be its change in volume if it is heated from 30˚Cto 200˚C. The coefficient of linear expansion of the quartz is 0.4x10ˉ⁶Kˉ¹.
Vol of sphere = 4/3 ∏r³
= 4/3∏r³ x (8.75x10ˉ²/2)
= 3.51 x 10ˉ⁴m³
V = Vo (1+3αΔT)

= 3.51 x 10ˉ⁴m³ (1+3(0.4x10ˉ⁶Kˉ¹.)(200-30)
= 3.511 x 10ˉ⁴m³
ΔV = Vf-Vo
= 3.511 x 10ˉ⁴m³ -3.51 x 10ˉ⁴m³
= 0.1 x 10ˉ⁶m³





6. A linear rod 40cm long at 40˚C is heated to 60˚C. The length of the rod is then measured to be 40.0105cm. what is the coefficient of linear expansion of the metal.

ΔV = βVo ΔT
ΔV = V-Vo
= 0.40105 – 0.4
= 1.05x10 ˉ⁴
β = 2α
β= (ΔV/ VoΔT)
= (1.05 x 10ˉ⁴)/(0.4 (60-20))
= 6.56 x 10ˉ⁶X2
= 13 x10ˉ⁶Kˉ¹





7. An iron ring is to fit snugly on a cylinder iron rod. At 20˚C, the diameter of the rod is 6.420cm. to slip over the rod diameter by about 0.008cm. to what temperature must the ring be brought if its hole is to be large enough so it will slip over the rod? Given α iron = 12 x 10ˉ⁶Kˉ¹


To = 20˚C
Lo = 6.420cm
ΔL= 0.008cm
αiron = 12 x 10ˉ⁶Kˉ¹

ΔT = ΔL / αiron Lo
= (0.008 x 10ˉ²)/ (12 x 10ˉ⁶(6.420 x 10ˉ²))
= 103.8˚C

ΔTF = T + To
= 103.8 + 20
= 123.8˚C





8. The confficient of linear expansion of aluminium is 23 x 10ˉ⁶Kˉ¹. a circular hole in an aluminium plate is 2.725cm in diameter at 0˚C. What is the diameter of the hole if the temperature of the plate is raised to 100˚C.

α= 23 x 10ˉ⁶Kˉ¹.
Vo = 2.725cm
To = 0˚C
TF = 100˚C
ΔV = Vo (1+3αΔT)
=2.725x 10ˉ² (1+3(23x10ˉ⁶Kˉ¹)(100-0)
= 2.7cm





9. The coefficient of linear expansion of steel is 12 x 10ˉ⁶Kˉ¹. A railroad track is made of invalidual rail of steel 1.0km in length . by what length wouldthese rails change between a cold day when the temperature is -10˚C and hot day at 30˚C.

αsteel= 12 x 10ˉ⁶Kˉ¹ Lo = 1000m To = -10˚C TF = 30˚C

ΔL = αsteel LoΔT
= 12 x 10ˉ⁶Kˉ¹(1000m)(30-(-10))
= 48x 10 ˉ²
= 48cm

10. A cooper plate a length of 0.12m and a width of 0.10m at 25˚C. The plate is uniformly heated to 175˚C. If the linear expansion coefficient for cooper is 1.70 x 10ˉ⁵Kˉ¹ . what is the change in the area plate as a result of the increase in temperature?


Ao = 0.12cm
To = 25˚C
TF = 175˚C
α= 1.70 x 10ˉ⁵Kˉ¹

ΔA = 2αAoΔT
= 2(1.70 x 10ˉ⁵Kˉ¹)(0.12)(175-25)
=6.12 x 10ˉ⁵m²