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?

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained,

Heat loss at rate, q = 4.68 kJ/min

Resistance, R =

Operating Voltage,

Now,

Power that the resistor releases,

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

t = 0.3852

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