Question

Temperature of a tungsten filament of 90watt electric bulb is 2000K Fond surface area of filament.
Er= 0.3

Answer 1

Given:-

• Temperature = T = 2000K
• Power = 60 watt
• er = 0.3

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Need to find:-

• Surface area of filament = ?

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Need to know:-

• $E = \dfrac{Q}{AT}$

• $E= \sigma \: {T}^{4} \times er$

• $P = \dfrac{Q}{T}$

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Solution:-

We know,

$E = \dfrac{Q}{AT}$

Also,

$E= \sigma \: {T}^{4} \times Er$

Thus,

$\dfrac{Q}{AT} = \sigma \: {T}^{4} \times 0.3$

$\implies \dfrac{P}{A} = 5.67 \times {10}^{ - 8} \times 16 \times {10}^{12} \times 0.3$

$\implies A = \dfrac{600}{5.67 \times 16 \times {10}^{4} \times 0.3}$

$\implies A = \dfrac{150}{22.68} \times {10}^{ - 4}$

$\red{\bold{\boxed{\large{\implies A = 15 \times {10}^{ - 4}}}}}$

Area is $15 \times {10}^{ - 4}\: m^{2}$

Answer 2

just evaluate by multiplying 0.3 to 20£

means you will get 60,000

change it to celcius

after that

you will get 60° C

hope this helps

thanks