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 the time is 38.67%.

Step-by-step explanation:

Given that,

Energy = 4.68 KJ

Resistance = 60

Voltage =110 V

If the rate of heat energy supplied by the coil to the oil bath = Q

`Q=4.68\ kJ/min`

We need to calculate the power released by the resistor at voltage

`P=(V^2)/(R)`

Put the value into the formula

`P=(110^2)/(60)`

`P=201.7\ W`

`P=12.102\ KJ/min`

We need to calculate the fraction of the time

`T=(Q)/(P)`

Put the value into the formula

`T=(4.68)/(12.102)`

`T=0.3867`

The percentage of time is

`T = 38.67\%`

Hence, The fraction of the time is 38.67%.