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Standard XII

Physics

Convection

A continuous ...

Question

A continuous flow water heater (geyser) has an electrical power rating $= 2 kW$ and efficiency of conversion of electrical power into heat $= 80 %$ . If water is flowing through the device at the rate of $100 cc/sec$ , and the inlet temperature is $10^{\circ} C$ , the outlet temperature will be

{12.2}^{\circ} C

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{13.8}^{\circ} C

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20^{\circ} C

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{16.5}^{\circ} C

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Solution

The correct option is B ${13.8}^{\circ} C$
For 1 sec we can say that
$P_{c} \times 80 % = (ρ v) s (t - 10)$
$2 \times 10^{3} \times \frac{80}{100} = (1000) .100 \times (10^{- 2})^{3} . 4200 (t - 10)$
On solving
$t = {13.8}^{\circ} C$

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A continuous flow water heater (geyser) has an electrical power rating =2 kW and efficiency of conversion of electrical power into heat =80%. If water is flowing through the device at the rate of 100 cc/sec, and the inlet temperature is 10∘C, the outlet temperature will be

The correct option is B 13.8∘CFor 1 sec we can say that Pc×80%=(ρv)s(t−10) 2×103×80100=(1000).100×(10−2)3.4200(t−10) On solving t=13.8∘C

A continuous flow water heater (geyser) has an electrical power rating $= 2 kW$ and efficiency of conversion of electrical power into heat $= 80 %$ . If water is flowing through the device at the rate of $100 cc/sec$ , and the inlet temperature is $10^{\circ} C$ , the outlet temperature will be

The correct option is B ${13.8}^{\circ} C$
For 1 sec we can say that
$P_{c} \times 80 % = (ρ v) s (t - 10)$
$2 \times 10^{3} \times \frac{80}{100} = (1000) .100 \times (10^{- 2})^{3} . 4200 (t - 10)$
On solving
$t = {13.8}^{\circ} C$