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2 years ago

I have 2 bulbs one coverts 60% of energy to light 45% which is the most efficient and what happens to the rest of the energy

Physics

2 answers:

Leya [2.2K]2 years ago

5 0

Answer:

The first one (60%)

Explanation:

The first one converts 15% more energy than the other one. Therefore, it is more efficient.

Hope this helps!

Hitman42 [59]2 years ago

4 0

Answer:60% one is the most efficient one

Explanation: First one is given 60% of light energy out of 100% and second one is given 45% light energy out of 100%. Waste of first one is 40% and second one is 55% .

Then those waste energy is transfer to the heat energy.

When you touch the bulb you feel hot . Its because of heat energy. If the heat loss is less . Bulb is more efficient

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A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

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