Question

A resistor held in running water carries electric current. Treat the resistor as the system

1. Does heat flow into the resistor?
2. Is there a flow of heat into the water?
3. Is any work done?
4. Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.

Answer 1

(a) The passage of electric current causes heat to be generated in the resistor. The heat generated = I₂Rt in standard notation.

(b) Yes. The resistor generates heat in the water.

(c) \[\ce{\underset{(Q)}{I^2Rt} = \underset{(\Delta U)}{MCΔT} + \underset{(W)}{P \Delta V}}\]

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, Δ V = increase in the volume of the water.

(d) There is no mention of the change in volume. So, there is no work done in this case.

∴ W = 0

First law of Thermodynamics,

ΔU = Q - W

The resistor is heated due to the joules heating effect. So, that it would transfer the heat to water. So, the amount of heat Q will be negative.

ΔU = - Q - 0

ΔU = - Q