Thermal Properties of Matter

JEE Main level, single correct, medium

82 of 126

Question 82 of 126

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27oC to 77oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

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Answer: (A) 8 min 20 s

P=1000watt\text{P}=1000\text{watt}

∴Pin=(1000−160)J/s=840J/s∴{\text{P}}_{\text{in}}=\left(1000-160\right)\text{J/s}=840\text{J/s}

Now, dQdt=msdTdt\dfrac{\text{d}\text{Q}}{\text{dt}}=\text{ms}\dfrac{\text{dT}}{\text{dt}}

∴840=msdTdt∴840=\text{ms}\dfrac{\text{dT}}{\text{dt}}

∴dt=ms840dT∴\text{dt}=\dfrac{\text{ms}}{840}\text{dT}

Integrating both sides we have;

t=ms840∫dT\text{t}=\dfrac{\text{ms}}{840}\int \text{dT}

=(2×1000×10−6×1000)×4.2×1000840×50=\dfrac{\left(2\times 1000\times 1{0}^{-6}\times 1000\right)\times \text{4.2}\times 1000}{840}\times 50

=500sec=500sec

=8min20sec=8min20sec

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