Question

An electric iron has a power rating of 750W

a. How many joules of electric energy does it change into heat energy every second?

b. How many joules of work can it do in 3 seconds

c. How long does it take the iron to do 1500J of work?​

Answer 1

a) P = E / T

or every second it means after 1 sec

750 = E / 1

E = 750 / 1 = 750 J = 0.75 KJ

b ) P = E / T

E = P X T

E = 750 x 3 = 2250 J = 2.250 KJ

work done in 3 sec is 2.250 KJ.

c) E = 1500J

P = 750 watt

hence by formula , P = E / T

T = E / P = 1500 / 750 = 2 sec

time taken for 1500 J is 2 sec....

hoped so u got the ans mate ...thnks me later