2
(
)
H = (3A) à 20⌠à 120 s
Step II
H = 21 600J
To find the energy (unit consumed) kWh
Step 2: To find heat gained by water and
calorimeter
ííŽíŚíííŤ í¨í íŽí§í˘ííŹ íŽíŹíí
(Energy consumed )
[
]
íť = íę˘íśę˘âí + íśíâθ
Power
(in kW)
Time
(in hours)
= [
] Ă [
]
íť = 0.1 Ă 4200âí + 40âí
íť = 460âí
[
]
[
]
= 0.3kW Ă 3.5h
= 1.05kWh
Power consumed by five bulbs in each day
(in kWh ) = 1.05kWh
Step 3: To find the change in temperature,
âí
íťíííĄ í í˘íííííí
â
â
ííŚ íííę˘íĄííę˘ ę˘í˘íííííĄ
Step III
íťíííĄ íííííí ííŚ í¤ííĄíí
= |
|
To find the total energy in 20 days (unit
consumed) kWh
ííí ę˘ííííííííĄíí
[
]
[
]
í¸ = 1.05kWh Ă 20
21 600 = 460âí
21600
âí =
= 21kWh
460
âí = 46.96â
Step IV
To find the total cost in 20 days
âí â 47â
íí§ííŤí í˛
íí¨íŹí íŠííŤ
íŽí§í˘í
The temperature rise â 47â
Total cost = [
] Ă â
â
(in kWh)
Example 08
= 21 Ă 150Tsh.
(a) If an electric cattle contains 960W
heating element (unit), what current
does it take from a 240V mains?
= Tsh. 3,150
Hence, the cost of watching this TV
for 30 days is Tsh. 3,150
P = IV
í
I =
íź
Example 07
A current of 3A flow for 20min through a
wire of resistance 20âŚ. If the wire is totally
immersed inside a 0.1kg of water in a can of
960
I =
240
heat
capacity
40J/K.
Calculate,
the
I = 4í´
temperature rise of water. The specific heat
capacity of water = 4200J/kgK for water.
(b) (i) How long will the kettle take to rise
2kg of water at 15â to the boiling
point, if 90% of the heat produced is
used raising the temperature of water ?
Solution
Step 1: To find heat supplied by electric
current
Solution
í = íííí
Step 1: Heat supplied by electric current