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Standard XII

Chemistry

Enthalpy of Combustion

A kettle with...

Question

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

8 min 20 s

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Solution

The correct option is A 8 min 20 s
Volume, V = 2 litre = $2 \times 10^{- 3} m^{3}$
Energy supplied by heater = $1000 J s^{- 1}$
Heat lost to the atmosphere = $160 J s^{- 1}$
Heat absorbed by water = $1000 - 160 = 840 J s^{- 1}$
$\frac{q}{t} = \frac{ms Δ T}{t} = \frac{V ρ S Δ T}{t} 840 = \frac{2 \times 10^{- 3} \times 10^{3} \times 4200 \times 50}{t} \Rightarrow t = 500 s = 8 min 20 s$

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

The correct option is A 8 min 20 sVolume, V = 2 litre = 2×10−3m3 Energy supplied by heater = 1000 Js−1 Heat lost to the atmosphere = 160 Js−1 Heat absorbed by water = 1000−160=840 Js−1 qt=msΔTt=VρSΔTt840=2×10−3×103×4200×50t⇒t=500 s = 8 min 20s