The difference in length between a brass and an iron is 14CM at 10°C what must be the length of the iron for this difference to remain 14CM when both rods are heated to 100°C? Given that Linear expansivity of brass is 19 × 10^-6/K and lenear expansivity of iron is 12×10^-6/K
*Solution✍🏼*😄
Let length of brass be *a*
Length of iron be *b*
All length above are at 10°c
✍🏼Their differences is 14cm
a-b=14...(i)
*In equation (i) above is (a-b) because brass expands faster than iron*😊😊
✍🏼At 100°c means change in temperature is 90°c
From ΔL=αΔθ×L
For brass
ΔL=α1Δθ×a.....(ii)
For iron
ΔL=α1Δθ×b....(iii)
Take equation (ii)=(iii) Now🥸 why?
*be cause difference in length to remain the same*
αΔθ×b=αΔθ×a
19×10-⁶a×90°=12×10-⁶×90°×b
19a=12b
a=(12/19)b.....(iv)
Take equation (iv) into equation (i)
a-b=14
(12/19)b-b=14
12b-19b=266
'7b=266
divide by 7 both sides u get
b=-38cm correct answer but negative
Therefore the length of iron is 38cm
Thanks alot for that solution .but why the answer bring us the negative sign
The answer is negative because when the temperature apply, it should be the length to increase but the length remain unchanged so it must occur negative length to show that the length is oppose the reality
L1=L2
I need the solution
About the above questions
The difference in length between a brass and an iron is 14CM at 10°C what must be the length of the iron for this difference to remain 14CM when both rods are heated to 100°C? Given that Linear expansivity of brass is 19 × 10^-6/K and lenear expansivity of iron is 12×10^-6/K
To find the length of the iron when both rods are heated to 100°C while maintaining a 14cm difference in length, we can use the concept of linear expansion.
First, let's calculate the initial length difference between the brass and iron rods at 10°C. We are given that the initial difference is 14cm.
Next, we need to consider the linear expansivity of both brass and iron. The linear expansivity is given as follows:
- Brass: 19 × 10^-6/K
- Iron: 12 × 10^-6/K
Since we are dealing with a change in temperature from 10°C to 100°C, the temperature change is ΔT = 100°C - 10°C = 90°C.
Now, let's calculate the change in length for both brass and iron rods separately using their respective linear expansivity values:
For brass: ΔL_brass = L_brass * α_brass * ΔT, where ΔL_brass is the change in length of brass, L_brass is the initial length of brass, α_brass is the linear expansivity of brass, and ΔT is the temperature change.
For iron: ΔL_iron = L_iron * α_iron * ΔT, where ΔL_iron is the change in length of iron, L_iron is the initial length of iron, α_iron is the linear expansivity of iron, and ΔT is the temperature change.
Since we want the difference between brass and iron to remain constant, we can set ΔL_brass equal to ΔL_iron, considering that they both have the same temperature change (ΔT = 90°C).
ΔL_brass = ΔL_iron
L_brass * α_brass * ΔT = L_iron * α_iron * ΔT
Dividing both sides of the equation by ΔT, we get:
L_brass * α_brass = L_iron * α_iron
Now, we can substitute the values we know:
L_brass * (19 × 10^-6) = L_iron * (12 × 10^-6)
Simplifying the equation:
L_brass / L_iron = (12 × 10^-6) / (19 × 10^-6)
Since we know that L_brass - L_iron = 14cm initially, we can use this to find L_brass in terms of L_iron:
L_brass = L_iron + 14cm
Substituting this into the equation:
(L_iron + 14cm) / L_iron = (12 × 10^-6) / (19 × 10^-6)
Now, we can solve for L_iron:
(L_iron + 14cm) = (12 × 10^-6) / (19 × 10^-6) * L_iron
Simplifying further:
L_iron + 14cm = 0.632*L_iron
Rearranging the equation:
0.368*L_iron = 14cm
Solving for L_iron:
L_iron = 14cm / 0.368
Calculating L_iron:
L_iron ≈ 38.04 cm
Therefore, the length of the iron rod when both rods are heated to 100°C while maintaining a 14cm difference is approximately 38.04 cm.