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Answer the following questions · Q2

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

(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
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Step 1. (a) The resistor is not receiving heat from an outside heat source — instead, it GENERATES heat internally due to Joule heating from the current passing through it, so no heat flows INTO it from outside.

Step 2. (b) This internally-generated heat is then carried away by the surrounding running water, which is cooler, so heat DOES flow out of the resistor into the water.

Step 3. (c) Electrical work is done ON the resistor by the source (battery/supply) driving the current through it — treating the resistor as the system, this is work being done ON the system, so WW (work done BY the resistor) is negative.

Step 4. (d) Since the resistor's state (its resistance, hence its internal energy/temperature) is unchanged in this steady-state situation, ΔU=0\Delta U = 0. From the first law, ΔU=Q−W=0⇒Q=W\Delta U = Q - W = 0 \Rightarrow Q = W. Since WW is negative (work done on the resistor), QQ is also negative — meaning the resistor rejects heat, and in fact rejects heat equal in magnitude to the electrical work supplied to it: all the electrical energy is dissipated as heat into the water.

✓Final answer

  1. No — heat doesn't flow into the resistor from outside; it's generated inside it.
  2. Yes — heat flows out of the resistor into the water.
  3. Yes — electrical work is done ON the resistor.
  4. ΔU = 0 (steady state) ⇒ Q = W: all the electrical work supplied is rejected as heat to the water.

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