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GarryVolchara [31]
3 years ago
13

An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an

electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained, T_{o} = 50.5^{\circ}

Heat loss at rate, q = 4.68 kJ/min

Resistance, R = 60\Omega

Operating Voltage, V_{o} = 110 V

Now,

Power that the resistor releases, P_{R} = \frac{V_{o}^{2}}{R}

P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min

The fraction of time for the current to be turned on:

P_{R} = \frac{q}{t}

12.148 = \frac{4.68}{t}

t = 0.3852

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