Water of volume 2 L in a closed container is heated with a coil of...

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 27^oC to 77^oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Options

A8 min 20 s
B7 min
C14 min
D6 min 2 s

Correct answer

A. 8 min 20 s

Step-by-step solution

\text{P} = 1 0 0 0 \text{watt}

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

Now,

\dfrac{\text{d} \text{Q}}{\text{dt}} = \text{ms} \dfrac{\text{dT}}{\text{dt}}

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

∴ \text{dt} = \dfrac{\text{ms}}{8 4 0} \text{dT}

Integrating both sides we have;

\text{t} = \dfrac{\text{ms}}{8 4 0} ∫ \text{dT}

= \dfrac{\left(2 \times 1 0 0 0 \times 1 {0}^{- 6} \times 1 0 0 0\right) \times \text{4.2} \times 1 0 0 0}{8 4 0} \times 5 0

= 5 0 0 sec

= 8 min 2 0 sec

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