JEE PYQ: Thermodynamics Question 36
Question 36 - 2019 (12 Apr Shift 2)
One kg of water, at 20$°$C, is 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$°$C), Latent heat of water = 2260 kJ/kg]
(1) 16 minutes
(2) 22 minutes
(3) 3 minutes
(4) 3 minutes
Show Answer
Answer: (2)
Solution
Power $= \frac{V^2}{R} = \frac{200^2}{20} = 2000$ W
Heat required = heat to raise temperature + heat to vaporize
$= 1 \times 4200 \times (100 - 20) + 1 \times 2260 \times 10^3$
$= 336000 + 2260000 = 2596000$ J
Time $= \frac{2596000}{2000} = 1298$ s $\approx 22$ minutes
BETA
