One $k g$ of water, at $20^{o} C$ , is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20 Ω$ . The rms voltage in the mains is $200 V$ . Ignoring heat loss from the kettle. time taken for water to evaporate folly. is close to:
[Specific heat of water $= 4200 J / k g^{o} C$ ), Latent heat of water $= 2260 k J / k g$ ]

3

minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22

minutes

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10

minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $22$ minutes
$Q = P \times t$
$Q = m c Δ T + m L$
$P = \frac{V_{r m s}^{2}}{R}$
$4200 \times 80 + 2260 \times 10^{3} = \frac{(200)^{2}}{20} \times t$
$t = 1298 s e c$
$t ≃ 22 m i n$

Suggest Corrections

Similar questions

Q. It takes 15 minutes for an electrical kettle to heat a certain quantity of water in grams from

0^{o} C

to boiling point

100^{o} C

. If it takes 81 minutes to boil water into steam, the latent heat of steam is :

An electric kettle raises the temperature of the certain mass of water from 0 $^{\circ} C$ to 100 $^{\circ} C$ in 2.1 minutes. It takes 11.2 minutes to boil off the water. Calculate the specific heat of vaporization.

Q. An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes.If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be :

Q. The water in an electric kettle begins to boil in ten minutes after being switched on. The wire used as the heating element is modified so that the water begins to boil in 7 minutes. How can this be achieved?

A student boils the water in an electric kettle for 20 minutes. Using the same main supply, he wants to reduce the boiling time of water. To do so, should he increase or decrease the length of the heating element? Justify your answer

Join BYJU'S Learning Program

The correct option is B 22 minutesQ=P×tQ=mcΔT+mLP=V2rmsR4200×80+2260×103=(200)220×tt=1298sect≃22min

The correct option is B $22$ minutes
$Q = P \times t$
$Q = m c Δ T + m L$
$P = \frac{V_{r m s}^{2}}{R}$
$4200 \times 80 + 2260 \times 10^{3} = \frac{(200)^{2}}{20} \times t$
$t = 1298 s e c$
$t ≃ 22 m i n$