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Specific heat capacity

One kilogram of water at 20^{} {C} is heated in an electric ke

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Solving time: 3 mins

One kilogram of water at $2 0^{\circ} C$ is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20Ω$ . 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 $= 4200 J / (kg^{\circ} C)$ , Latent heat of water $= 2260 kJ / kg$ ] (a) $16 min$ (b) $22 min$ (c) $3 min$ (d) $10 min$

Views: 5,839 students

Updated on: Feb 10, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

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Alexander Discussed

One kilogram of water at

2 0^{\circ} C

is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of

20Ω

. 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^{\circ} C)

, Latent heat of water

= 2260 kJ / kg

] (a)

16 min

(b)

22 min

(c)

3 min

(d)

10 min

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Exp. (b)
Heat required by water for getting hot and then evaporated is
Here,

m = 1 kg, Δ T = 10 0^{\circ} - 2 0^{\circ} = 8 0^{\circ} C

s = 4200 J kg^{- 1}^{\circ} C^{- 1}, L = 2260 \times 1 0^{3} J kg^{- 1}

So, heat required is

Δ Q = 1 \times 4200 \times 80 + 1 \times 2260 \times 1 0^{3}

= 336 \times 1 0^{3} + 2260 \times 1 0^{3}

= 2596 \times 1 0^{3} J

This heat is provided by a heating coil of resistance

R = 20Ω

connected with

AC

mains

V_{r m s} = 200 V

So, heat supplied by heater coil is

Q = Pt = \frac{V _{mi}^{2}}{R} \times t

where,

P =

power and

t =

time.

= \frac{( 200 ) ^{2}}{20} \times t

= 2 \times 1 0^{3} \times t

Substituting the value of heat from Eq. (ii), we get

t = \frac{2956 \times 1 0 ^{3}}{2 \times 1 0 ^{3}} s = \frac{2956}{2 \times 60} min = 24.63 min

Nearest answer is 22 min.

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Question Text	One kilogram of water at $2 0^{\circ} C$ is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20Ω$ . 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 $= 4200 J / (kg^{\circ} C)$ , Latent heat of water $= 2260 kJ / kg$ ] (a) $16 min$ (b) $22 min$ (c) $3 min$ (d) $10 min$
Updated On	Feb 10, 2023
Topic	Thermodynamics
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 4
Upvotes	548
Avg. Video Duration	8 min

Text solutionVerified

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