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Standard X

Physics

Calorimeter

A kettle with...

Question

A kettle with $2$ litre water at $27^{\circ}$ C is heated by operating coil heater of power $1$ kW. The heat is lost to the atmosphere at constant rate $160$ J/sec, when its lid is open. In how much time will water be heated to $77^{\circ}$ C with the lid open'? (specific heat of water = $4.2$ kJ/k g)

8 min 20 sec

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Solution

The correct option is A 8 min 20 sec
By the law of conservation of energy. energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.
$P t = m s$ $Δ θ$ + energy lost
$1000 t = 2 \times 4.2 \times 10^{3} \times 50 + 160 t$
$840 t = 8.4 \times 10^{3} \times 50 = 500 s e c = 8 m i n 20 s e c$

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A kettle with 2 litre water at 27∘C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J/sec, when its lid is open. In how much time will water be heated to 77∘C with the lid open'? (specific heat of water = 4.2 kJ/k g)

The correct option is A 8 min 20 secBy the law of conservation of energy. energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.Pt=ms Δθ + energy lost1000t=2×4.2×103×50+160t840t=8.4×103×50=500sec=8min 20sec