Question

A cylindrical rod having temperature t1 and t2 at its end. the rate of flow of heat q1 cal s-1. if all the dimensions (length and radius) are doubled keeping temperature constant,then the rate of flow of heat q2 will be

Answer 1

Hey mate,

# Answer- Q2=2Q1

# Explaination-

Rate of heat flow is given by

Q1=H/t

Q1=KA(θ1−θ2)/l

Now all linear dimensions are doubled,

length=2l

Area=4A ...coz radius is doubled

Q2=K(4A)(θ1−θ2)/(2l)

Q2=2KA(θ1−θ2)/l

Q2=Q1

Rate of flow of heat Q2=2Q1.

Hope this helps you.

Keep asking...

Answer 2

Answer:

q₂ = 2q₁

Step-by-step explanation:

Required formula: Q/t = K.A.ΔT/l

Q/t = Rate of heat flow.

K = Heat transfer coefficient.

ΔT = Difference in temperature.

l = Length.

Given, for condition 1; Q/t = q₁

and for condition 2; Q/t = q₂

ΔT = t₂ - t₁ is same in both conditions.

Let's assume the radius and length of the rod in condition 1 is 'r' and 'L' respectively.

Thus, in condition 2; radius and length are 2r and 2L.

Thus, area of heat flow in condition 1 = π.r²

Thus, area of heat flow in condition 2 = π.(2r)² = 4πr²

q₁ = K.(πr²).ΔT/L

q₂ = K.(4πr²).ΔT/2L = 2[ K.(πr²).ΔT/L] = 2q₁