# Heat Developed in a Resistor

What does the graph that represents the total heat developed from time t = 0 by a resistor carrying a steady current look like (where heat is on the y-axis and time is on the x-axis)?

The answer given is increasing from the origin with a constant slope. I thought it would be constant.

I know that heat developed in a current carrying wire is proportional to the resistance. In a resistor, with steady current, the amount of resistance remains constant, which is what led me to believe that the graph would be a horizontal line. Also, we do not see our appliances get hotter and hotter indefinitely, which I thought would eliminate the graph starting at the origin and increasing linearly (which turned out to be the answer).

What is the flaw in my understanding?

• You say "heat", but do you really mean temperature? Jun 17, 2016 at 15:04

The resistor constantly produces heat with rate of $IR^2$; however, not all of this energy remains in the resistor (in form of internal thermal energy, ie. temperature).

So there's a sink: heat exchange between the resistor and environment (by radiation, conduction and convection).

Total change in temperature relates to net effect of the two. If the resistor constantly produces heat with some rate, but it also radiates energy with the same rate to the environment, the temperature basically remains constant, and this is what happens when you let a resistor go for a while.

EDIT (More Explanation):

Let's see what happens to a light bulb when you turn it up: Lamps are resistors (incandescent lamps).

• When a lamp is off, heat production rate due to electricity is 0, and heat radiation rate is 0 too.

• Just after you light it up, it begins to produce electrical energy with a rate of, say $100$ $J/s$ while heat radiation is still low, say $10$ $J/s$ because the lamp is still cold. This means that every second, $90$ $J$ of energy becomes internal energy and raises temperature of the lamp.

• After a while, lamp gets warm. Warm objects release heat faster, say $60$ $J/s$. But it's still gaining heat with the $100$ $J/s$ rate. Therefore it's still warming up, but a bit slower; $40$ $J/s$.

• When lamp gets hot enough so that it radiates energy with $100$ $J/s$ rate, since this is equal to electrical heat produced, these factors cancel out, building a balanced condition called equilibrium. At this moment, temperature no longer changes; even tough the heat is being produced all the time.

The RATE at which the resistor is heating is constant. But the total amount of heat accumulated over time is equal to the RATE times TIME, just like in a rate-time-distance problem.

Rate of heat production = power = current$^2 \times$ resistance = $I^2R =$ constant $P$

Heat produced = power $\times$ time $\Rightarrow$ $H = Pt$

This is of the form of the equation of a straight line $y=mx+c$

So a graph of $H$ against $t$ should be a straight line through the origin and of slope $P$.

Heat energy in a circuit is directly proportional to time, so a heat - time graph would be a straight line passing through the origin with a certain slope(power). The rate at which heat is being emitted by the resistor is constant(the slope of the graph or power). You are only wrong in assuming that the heat emitted by the resistor stays in the resistor while in reality the resistor loses heat to the environment via processes like conduction and radiation. The rate at which the resistor is heated and the rate at which it releases that heat is equal. If the two weren't equal, the resistor would keep on heating and would eventually melt down or the resistor would keep on cooling until it reached -273 degrees celsius(absolute zero).