Question

3.if an electric iron of 1200W is used for 20 mins everyday find the electric energy consumed in
the month of September.
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Answer 1

Answer:

hello mate

for refrigerator

convert 400 w into Kw by dividing it by thousand.

4000/1000 = .4 Kw

energy consumed in the 30days = .4 * 8 * 30 = 96 KwH  

for iron

convert 750 w into Kw by dividing it by thousand.

750/1000 = 3/4 Kw

energy consumed in the 30days = 3/4 * 2 * 30 = 45 KwH

total energy consumed = 96+45 = 141

cost of 1kwh = 3rs

so 141 kwh = 3 × 141 = 423rs

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Answer 2

Given:-

• Power of Electric Iron ,P = 1200 W

• Time ,t = 20 min = 20/60 = 1/3 hour

• No. of Days in Month of September = 30

To Find:-

• Energy Consumed ,E

Solution:-

We have to calculate the energy consumed by iron in the month of September .

Energy is defined as the product of Power and time .

• E = P×t

where,

E is the Energy Consumed

P is the power of Iron

t is the time taken

Substitute the value we get

→ E = 1200 × 30 × 1/3

→ E = 1200 × 10

→ E = 12000 J

→ E = 12000/1000 KJ

→ E = 12 KJ

Therefore,the energy consumed by the electric iron is 12 KiloJoules .