Two solid cylindrical copper rods, P and Q, are maintained under identical temperature dif… — JAMB UTME

Two solid cylindrical copper rods, $P$ and $Q$ , are maintained under identical temperature differences across their ends. Rod $P$ has twice the radius and half the length of Rod $Q$ . What is the ratio of the rate of heat conduction through Rod $P$ to that through Rod $Q$ ?

$8 : 1$ Answer
B
$4 : 1$
C
$2 : 1$
D
$1 : 8$

Answer

The ratio of the rate of heat conduction through Rod

P

to that through Rod

Q

8 : 1

The rate of heat conduction is given by

\frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

. For Rod

P

, substituting

r_P = 2 r_Q

and

L_P = 0.5 L_Q

gives

\frac{(2)^2}{0.5} = 8

times the rate of Rod

Q

. Thus, the ratio of heat conduction rate is

8 : 1

Step-by-Step Solution

Write the fundamental equation for the rate of thermal conduction through a uniform conductor.

\frac{Q}{t} = \frac{k A \Delta T}{L}

Heat conduction rate is proportional to thermal conductivity

k

, cross-sectional area

A

, temperature difference

\Delta T

, and inversely proportional to length

L

Express the cross-sectional area

A

in terms of radius

r

for a cylindrical rod.

A = \pi r^2 \implies \frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

The cross-section of a cylindrical rod is a circle.

Set up the ratio of heat conduction rates for Rod

P

and Rod

Q

given

r_P = 2 r_Q

and

L_P = 0.5 L_Q

\frac{(Q/t)_P}{(Q/t)_Q} = \frac{r_P^2 / L_P}{r_Q^2 / L_Q} = \frac{(2 r_Q)^2 / (0.5 L_Q)}{r_Q^2 / L_Q} = \frac{4}{0.5} = 8

Material (

k

) and temperature difference (

\Delta T

) are identical for both rods and cancel out.

Key Concept

Thermal Conduction Rate Formula

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