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 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

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

Put the value into the formula

We need to calculate the fraction of the time

Put the value into the formula

The percentage of time is

Hence, The fraction of the time is 38.67%.

it is easily magnetized.

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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