Q. 10

# An electric kettl

Answer :

Given-

Time taken to boil 4 cups of water, *t* = 2 minutes

Volume of water boiled = 4 × 200 cc = 800 cc

Initial temperature, *θ*_{1} = 25°C

Final temperature, *θ*_{2} = 100°C

Change in temperature, *θ* = *θ*_{2} − *θ*_{1} = 75°C

Mass of water required to boil , *m* = 800 × 1 = 800 gm = 0.8 Kg

Now heat required for boiling water,

Where

m is the mass of the water

s is the specific heat of the water

θ is the change in temperature

Putting the value in the above formula, we get

We know, to convert watt-hour to watt-sec,

1000 watt – hour = 1000 × 3600 watt sec.

Hence cost of boiling 4 cups of water

(b)

Given,

Initial temperature, *θ*_{1} = 5°C

Final temperature, *θ*_{2} = 100°C

Change in temperature, *θ* = *θ*_{2} − *θ*_{1} = 95°C

heat required for boiling water,

Where

m is the mass of the water

s is the specific heat of the water

θ is the change in temperature

Putting the value in the above formula, we get

= 0.8 × 4200 × 95

= 319200

Again converting into watt-second,

Cost of boiling 4 cups of water

=11000×3600×319200

= Rs 0.09

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