Question

The heater element of a kettle has a constant resistance of 105ohm and apllied voltage of 240V
Calculate the time taken to raise the temperature of on litre of water from 15degrees celsius to 90 degrees celsius assuming that only 80% of the power input to the kettle is usefully employed
If the water equivalent of the kettle is 150g, find out how long it will take tobraise a second litre of water through the same temperature range immediately after the first time.

Answer 1

1) For the first time say time taken is t seconds. V = 240 volt; R = 105 ohm; mass of water m = 1000 gm; specific heat of water c = 4.186 J/gm , change in temperature = (90-15)°C = 75°C. 80% of the power input in the kettle is employed in heating the water.

2) Water equivalent of the kettle is 150 gm, i.e. the heat that is required to raise the temperature of the kettle by 1°C can be used to raise the temperature of 150 gm of water by 1°C.

That means total energy absorbed by the kettle+water system is actually usable energy and the rest is dissipated in the environment.

Total heat absorbed by the kettle+water system = total water equivalent mass * specific heat of water * change in temperature

= (150 + 1000) * 4.186 * 75 J

= 361042.5 J

When another 1 litre of water is heated immediately after the 1st 1 litre, the kettle is already at 90°C, therefore it does not absorb any heat. The whole absorbable heat of 361042.5 J is absorbed by the new 1 litre water.

Say this time it takes t1 seconds to reach from 15°C to 90°C.

(We are not considering efficiency as 80% here beause we have calculated the kettle+water system's total absorbable energy, which this time is being absorbed solely by the water since the kettle temperature is not changing)