I think I have an answer, but an not completely sure if it is correct, because first of all it’s been a long time I worked with formulaes, and secondly, I have some doubts about how I use Ohm’s law.

From what I understand the relationship is T(tip) = T(room) + kP, with:

T = Temperature (Kelvin)

P = power (Watt)

k = constant

Let’s define these first:

V = voltage (Volt)

I = current (Amp)

(d) = delta (“the change in”)

OK, the change in temperature of the tip is:

T(tip) - T(room) = (d)T = k(d)P

To get rid of that constant, we can say that the change in temperature is proportional to the change in power:

(d)T ~ (d)P

So, if the power is halved, the temperature is also halved. Like in the first question. Now, for the second question:

The power can be calculated using the following formula:

P = V * I

So, you would say that when V is halved, P is also halved, right? No, that’s not the case. We can’t say anything about P unless we know something about I. And the problem is, we don’t know anything about I.

Now comes the trick: Ohm’s law states:

V = I * R

which rewrites to:

I = V / R

We can substitute this into the formula for calculating the power:

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

(V^2 means “V square” or V * V)

Now we have a different formula for calculating P. But don’t we need to know something about R before we can say something about the change in P? Yes, and we know enough already:

Remember that a soldering iron generates heat becauses of high internal resistance. This value of R is so big, that it changes only little when varying V and I; therefore, this change can be neglected.

§P = §(V^2) / §R

becomes by neglecting the change in R:

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

which becomes

BP ~ (d)(V^2)[/B]

From this followes that when the voltage is halved, the power is quartered and with it the temperature of the tip.

Note that I’m not sure if I used Ohm’s law correctly here. Ohm’s law is used for calculating the resistance in an ideal conductor, and of course a soldering iron is far from ideal. However, since we don’t actually calculate the resistance here, I think it’s save to use it in this way.

Good luck with your exams.