Question

A heat flux of 4000 J/s is to be passed through a copper rod of length 10cm and area of cross-section 100cm2. The thermal conductivity of copper is 400 W/m°C. The two ends of this rod must be kept at a temperature difference of

(1) 1°C
(2) 10°C
(3) 100°C
(4) 1000°C

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Answer 1

Answer:

i hope it will help u alot

Answer 2

$\huge{\underline{\mathtt{\blue{Answer}}}}$

$\bold{Given}$

Heat flux [Q] = 4000 j/s

Length = 100cm²

The thermal conductivity [KA] = 400 W/m°C

We know the formula :

${\green{\overline{\green{\underline{\blue{\boxed{\pink{\mathtt{\frac{dQ}{dt} = \frac{KA∆T}{l}}}}}}}}}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = \frac{l}{K × A} × \frac{dQ}{dt}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T =\frac{0.1}{400 × 100 × 10^{4}} × 4000}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = \frac{1 × 10^{4} × 4000}{400 × 100 × 10}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = \frac{1 × 10000 × 4000}{400 × 100 × 10}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = \frac{40000000}{400000}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = \cancel\frac{40000000}{400000}}$

$\bold{\:\:\:\:\:\:\:\:\:\:∆T = 100°C}$

$\huge\bf\boxed{\boxed{\underline{\red{∆T = 100°C}}}}$