Answer:
1. i. 286.81 CAL ii. 1.2 KWH
2. 3150 J
Step-by-step 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