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Answered

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1) 1200 J is
the amount of energy that a clothes iron transforms per second, asks:
(I) How many CALs are there ?, (II)
How many KWH are

2) One
Low consumption electric lamp has an efficiency of 30% to illuminate.
If the electrical energy used is 4,500 J, determine the value of the
missing energies.

Sagot :

oladayoademosu

Answer:

1. i. 286.81 CAL ii. 1.2 KWH

2. 3150 J

Explanation:

1.  1200 J is  the amount of energy that a clothes iron transforms per second, asks:

i. (I) How many CALs are there ?

Since the amount of energy that a clothes iron transforms per second is 1200 J. We convert this to calories.

Since 1 cal = 4.184 J,

1200 J = 1200 J × 1 cal/4.184 J = 286.81 Cal.

(II)  How many KWH are there

Since the amount of energy that a clothes iron transforms per second is 1200 J = 1200 J/s = 1200 W = 1.2 kW.

So, the amount of energy consumed in 1 hour is 1.2 kW × 1 h = 1.2 kWh = 1.2 KWH

2) One  Low consumption electric lamp has an efficiency of 30% to illuminate.  If the electrical energy used is 4,500 J, determine the value of the  missing energies.

efficiency, η = Energy output E/Energy input E' × 100 %

= E/E' × 100%

Since η = 30% and E' = 4500 J,

E = ηE'/100 % = 30 % × 4500 J/100 % = 0.3 × 4500 J = 1350 J

The missing energy, ΔE = energy input - energy output = E' - E = 4500 J - 1350 J = 3150 J

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