Would the current of a circuit consisting of a power source and a resistor affect the rate at which the resistor can heat water? If so, why?

Asked by: Unknown

Answer

Yes. The DC power P dissipated by a electrical circuit is P = EI, where E is the voltage across
the circuit and I is the current through the circuit. In the case of a resistor, Ohm's law
applies, so that E = IR, where R is the resistance in ohms. Thus, the power dissipated by a
resistor with E volts impressed across it is P = I^2 R = (E^2)/R.

Heat, as used rigorously in physics, is simply another expression for energy, but applied to
so-called thermodynamic systems. Consequently, the units of heat are energy. Temperature (which
is often erroneously taken to be the same as heat) is a measure of the heat, or energy, content of
a substance, such as water. The more heat, or thermodynamic energy, contained by the substance,
the higher its temperature, and vice-versa. Temperature and heat are NOT identical, though one
says loosely that one 'heats' a substance when one really means that the temperature of the
substance is increased (by increasing the heat content)!

The greater the current I through the resistor, the greater the power P = I^2 R dissipated by the
resistor. Since power is the time rate of flow of energy (the units of power are energy per unit
time), the greater the power dissipated by the resistor, the more rapidly energy is dissipated by
the resistor.

If the resistor is immersed in a volume of water, the energy dissipated by the resistor is absorbed
by the water (ignoring minor amounts of energy radiated directly to space, etc.). That is, the
heat content of the water is increased by the amount of energy dissipated by the resistor. Since
temperature is a measure of heat content, the temperature of the water will, necessarily, increase.
Thus, the greater the current through the resistor, the more rapid will be the rise in the
temperature of the water.. Equivalently, the water would be 'heated' more rapidly.
Answered by: Warren Davis, Ph.D., President, Davis Associates, Inc., Newton, MA USA

'Physicists like to think that all you have to do is say, these are the conditions, now what happens next?'