JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 17)
One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature
averaged) resistance of 20 $$\Omega $$. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time
taken for water to evaporate fully, is close to :
[Specific heat of water = 4200 J/(kg oC), Latent heat of water = 2260 kJ/kg]
10 minutes
22 minutes
3 minutes
16 minutes
Explanation
P$$ \times $$t = mS$$\Delta $$t + mLv
$$ \Rightarrow $$ $${{{V^2}} \over R}t$$ = mS$$\Delta $$t + mLv
$$ \Rightarrow $$ $${{{{\left( {200} \right)}^2}} \over {20}}t$$ = $$1 \times 4200 \times \left( {100 - 20} \right)$$ + $$1 \times 2260 \times {10^3}$$
$$ \Rightarrow $$ 2000t = 336000 + 2260000
$$ \Rightarrow $$ t = 1298 s = 21.6 min $$ \simeq $$ 22 min
$$ \Rightarrow $$ $${{{V^2}} \over R}t$$ = mS$$\Delta $$t + mLv
$$ \Rightarrow $$ $${{{{\left( {200} \right)}^2}} \over {20}}t$$ = $$1 \times 4200 \times \left( {100 - 20} \right)$$ + $$1 \times 2260 \times {10^3}$$
$$ \Rightarrow $$ 2000t = 336000 + 2260000
$$ \Rightarrow $$ t = 1298 s = 21.6 min $$ \simeq $$ 22 min
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