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

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) = (V_(o)^(2))/(R)`

`P_(R) = (110^(2))/(60) = 201.67 W = 12.148 J/min`

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

`P_(R) = (q)/(t)`

`12.148 = (4.68)/(t)`

t = 0.3852