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Fed [463]
3 years ago
7

The kitchen in your house measures 10 ft by 10 ft. You need to have 50 footcandles of illumination in the room. How many 60 watt

bulbs will be required if one bulb emits 834 lumens. (1 footcandle = 1 lumen/sq.ft)
Physics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

<u>6 bulbs</u> are needed to illuminate the room.

Explanation:

Given:

Measurement of kitchen (A) = 10 ft by 10 ft = 100 sq. ft

Number of footcandles (n) = 50

Lumens emitted by 1 bulb = 834

Number of bulbs (N) = ?

We are also given,

1 foot candle = 1 lumen/sq. ft

So, 50 foot candles = 50 lumens/sq. ft

Now, for an area of 1 sq. ft 50 lumens are emitted.

So, for an area of 100 sq. ft, lumens emitted = 50 × 100 = 5000 lumens

Now, one bulb emits = 834 lumes

Therefore, number of bulbs required for emitting 5000 lumens is given as:

N=\frac{Number\ of\ lumens}{Number\ of\ lumens\ by\ 1\ bulb}\\\\N=\frac{5000}{834}=5.995\approx6

So, 6 bulbs are needed to illuminate the room.

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Answer:

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For isothermal expansion PV = constant

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Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

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So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

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1/0.19,

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P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

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T₃ = 90.1 K

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a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

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