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An electric immersion heater is switched on for 8 minutes. The heat supplied by it raises the temperature of 500 g of water from 10°C to 60°C. Calculate the power of the heater in watts.

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Question

An electric immersion heater is switched on for 8 minutes. The heat supplied by it raises the temperature of 500 g of water from 10°C to 60°C. Calculate the power of the heater in watts.

[Specific heat of water = 4.2 Jg−1°C−1]

Numerical
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Solution

Time = 8 min = 480 s

Let the power of the heater = P

Energy supplied by the heater = P × t = P × 480 s

Also, rise in temperature of water θR = (60 − 10) = 50°C

∴ Heat energy absorbed by water = mcθR

= 500 g × 4.2 Jg−1°C−1 × 50°C

= 105000 J

∴ By the law of conservation of energy,

P × 480 s = 105000 J

\[\therefore P =\frac{105000}{480}\]

= 218.75 J/s

= 218.75 W

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Chapter 10: Calorimetry - NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY [Page 238]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 10 Calorimetry
NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY | Q 4. | Page 238
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