Question 436: Thermodynamics & KTG

Question

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]

Accepted Answer

P$$ 	imes $$t = mS$$\Delta $$t + mL
v
$$ \Rightarrow $$ $${{{V^2}} \over R}t$$ = mS$$\Delta $$t + mL
v
$$ \Rightarrow $$ $${{{{\left( {200} ight)}^2}} \over {20}}t$$ = $$1 	imes 4200 	imes \left( {100 - 20} ight)$$ + $$1 	imes 2260 	imes {10^3}$$
$$ \Rightarrow $$ 2000t = 336000 + 2260000
$$ \Rightarrow $$ t = 1298 s = 21.6 min $$ \simeq $$ 22 min

Answer

10 minutes

Answer

22 minutes

Answer

3 minutes

Answer

16 minutes

Thermodynamics & KTG

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