ENERGY FROM THE SUN

The sun is the ultimate source of all forms of energy available on the earth. This can illustrated as follows :

WIND ENERGY:

Solar energy heats up the surface of the earth and the air near it. The hot air rises up and the cool air from above rushes to occupy its space. This makes the air to move. Moving air is known as wind and possesses kinetic energy. Thus, solar energy + air wind energy. Wind energy is converted into electrical energy in a wind farm using wind mills.

GREEN PLANTS MAKE THEIR FOOD:

Green leaves of plants make their food using sunlight (i.e. sun energy) by the process of photosynthesis. The cells of green leaves of plants contain chloroplasts. Each chloroplast contains chlorophyll (a green pigment) which converts carbon dioxide into sugar in the presence of sunlight by the process of photosynthesis. Process of photosynthesis is represented as follows :

Carbon dioxide + water Sugar+ Oxygen

The energy stored in the food is known as chemical energy. The food eaten by a man or an animal provides him the muscular energy, is used to do work. In other words, muscular energy is converted into mechanical energy. Thus,

Solarenergy  + Green leaves Food (chemical energy) Muscular energy Mechanical energy (work)

Question14. An electric bulb of 60W is used for 6 h per day. Calculate the ‘units’ of energy consumed in 1 day by the bulb.

Solution:  Power of bulb = 60 W

= 0.06 kW

Time used (t)  = 6 hr

Energy = power × time 

= 0.06 kW × 6h

= 0.36 kWh

= 0.36 units

Question 15. A family uses 250 units of electrical energy during a month. Calculate this electrical energy is Joules.

Solution:  1 kWh of Energy = 3.6 × 106 J

250 kWh of Energy = 3.6 × 106× 250 J 

= 9 × 108 J

Question16. A body does 20 joules of work in 5 sec. What is its power?

Solution:
 

      Power = 4 W.

Question 17. Derive relationship between kWh and Joule.

Solution:

1 kWh = 1 kW for 1 hour

1 kWh = 1000 W for 1 hour

1 watt=

1 kWh = 1000 for 1 hour.

1 kWh = 1000 seconds

1 kWh = 36,00,000 J

          = 3.6 × 106 J

Question18. A radio set of 60 watts runs for 50h. How many ‘units’ of electrical energy are consumed?

Solution:

 P = 60 watts

kW 

= 0.06 kW

          t = 50h

Energy = P × t

= 0.06 × 50 kWh

= 3 kWh

= 3 units.

Question19. An electric room heater with 2 rods is rated at 2kw. What is the cost of using it for 2 h a day for the month of September, if each unit costs Rs. 2.00?

Solution:

Power = 2 kW

t = 2 hours

Days = 30 days in September

Total time = 2 × 30 = 60 hr. 

= 120 kWh

Cost = 120 × 2.00 

        = Rs. 240