An immersion heater rated at 1000 W , 220 V is utilized to heat 0.01 m^3 of water. Assuming that the power is supplied at 220 V and 60\% of the supplied power goes into heating the water, determine how long it will take

Correct answer: B. 29 minutes Mass of water m = V × = 0.01 × 1000 = 10 kg Heat required to raise the temperature of water is given by Q = msΔ T Using the specific heat capacity of water s = 4200 J kg^ -1 ^ ^ ^ ^ -1 : Q = 10 × 4200 × (40 - 15) = 1050000 J Effective power utilized for heating P_ eff = 60\% of 1000 W = 600 W Time taken t = Q P_ eff = (1050000)/(600) = 1750 s Converting to minutes, t = (1750)/(60) 29.16 minutes 29 minutes

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Thermal and Chemical Effects of Electric Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Thermal and Chemical Effects of Electric Current

An immersion heater rated at 1000 W , 220 V is utilized to heat...

An immersion heater rated at

1000 \text{ W}

220 \text{ V}

is utilized to heat

0.01 \text{ m}^3

of water. Assuming that the power is supplied at

220 \text{ V}

and

60\%

of the supplied power goes into heating the water, determine how long it will take to raise the temperature of the water from

15^{}^{{}^{\circ}} \text{C}

40^{}^{{}^{\circ}} \text{C}

Options

A $17.5 \text{ minutes}$
B $29 \text{ minutes}$
C $10.5 \text{ minutes}$
D $44 \text{ minutes}$

Correct answer

B. $29 \text{ minutes}$

Step-by-step solution

Mass of water

m = V \times \rho = 0.01 \times 1000 = 10 \text{ kg}

Heat required to raise the temperature of water is given by

Q = ms\Delta T

Using the specific heat capacity of water

s = 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

Q = 10 \times 4200 \times (40 - 15) = 1050000 \text{ J}

Effective power utilized for heating

P_{eff} = 60\% \text{ of } 1000 \text{ W} = 600 \text{ W}

Time taken

t = \dfrac{Q}{P_{eff}} = \dfrac{1050000}{600} = 1750 \text{ s}

Converting to minutes,

t = \dfrac{1750}{60} \approx 29.16 \text{ minutes} \approx 29 \text{ minutes}

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