Question

2) A steel is 40 cm long at 20°C. The Coefficient of Linear expansion for steel is
12 x 10- (1/°C). What will be the change in length?
3) A 0.400 kg brick of copper is heated from 295k to 325k. How much jeat is absorbed by the iron?

i need this urgent pls​

Answer 1

2) :

The change in temperature (ΔT) = 70oC – 20oC = 50oC

The original length (L1) = 40 cm

Coefficient of linear expansion for steel (α) = 12 x 10-6 (Co)-1

Wanted : The change in length (ΔL) and the final length (L2)

Solution :

a) The change in length (ΔL)

Solution :

The change in length (ΔL)

ΔL = α L1 ΔT

ΔL = (12×10-6 oC-1)(40 cm)(50oC)

ΔL = (10-6)(24 x 103) cm

ΔL = 24 x 10-3 cm

ΔL = 24 / 103 cm

ΔL = 24 / 1000 cm

ΔL = 0.024 cm

b) The final length (L2)

L2 = L1 + ΔL

L2 = 40 cm + 0.024 cm

L2 = 40.024 cm

___________________

Answer 2

Temperature difference (T) = 70oC - 20oC = 50oC

L1 = 40 cm (original length).

Steel's linear expansion coefficient () = 12 x 10-6 (Co)-1

The length change (L) and the final length are both required (L2)

Alternative:

a) The length change (L)

The length change (L) is the solution.

T = L1 L = L1

(40 cm) L = (1210-6 oC-1) (50oC)

(10-6)(24 x 103) cm L =

24 x 10-3 cm = L

103 cm L = 24 cm

/ 1000 cm L = 24

0.024 cm = L

b) The completed length (L2)

L1 + L = L2

40 cm + 0.024 cm L2

40.024 cm L2