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Question
A resistor held in running water carries electric current. Treat the resistor as the system
- Does heat flow into the resistor?
- Is there a flow of heat into the water?
- Is any work done?
- Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Solution
(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
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