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 the time is 38.67%.

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

$P=\dfrac{V^2}{R}$

Put the value into the formula

$P=\dfrac{110^2}{60}$

$P=201.7\ W$

$P=12.102\ KJ/min$

We need to calculate the fraction of the time

$T=\dfrac{Q}{P}$

Put the value into the formula

$T=\dfrac{4.68}{12.102}$

$T=0.3867$

The percentage of time is

$T = 38.67\%$

Hence, The fraction of the time is 38.67%.