Question

A1 kW electric kettle contains 2 litre water at 27°C. The kettle is operated for 10 minutes. If heat is lost to the surrounding at a constant rate
of 160 J/sec, the temperature attained by water in 10 minutes will be (specific heat of water = 4.2kJ/kg°C)
(A) 57°C
(B) 67°C
ws
(C) 77°C
(D) 87°C

Answer 1

The temperature attained by water after 10 minutes is 87°C

Therefore, option (D) is correct.

Explanation:

Power of electric kettle = 1 kW = 1000 J/s

The heat lost to the surrounding is 160 J/s

Net heat transferred

$=1000-160$

$=840$ J/s

In 10 minutes the heat transferred by the electric kettle

$=10\times 60\times 840$ J

$=504000$ J

$=504$ kJ

Weight of 2 litre of water = 2000 gram = 2kg    (density of water = 1000 gram/litre)

Specific Heat of water $c=4.2$ kJ/kg°C

Initial temperature of water = 27°C

If the final temperature is T

Then

Using

$Q=mc\Delta T$

$504=2\times 4.2\times (T-27)$

$\implies T-27=\frac{504}{2\times 4.2}$

$\implies T-27=60$

$\implies T=27+60=87$°C

Hope this answer is helpful.

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