Question

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?

Answer 1

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