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

Answer: (a) No (b) Yes (c) Yes, electrical work is done on the resistor (d) \(\Delta U = 0\), so the heat given to the water equals the electrical work done on the resistor.

Take the resistor as the system. Its temperature stays steady because the running water keeps cooling it.

• (a) No heat flows into the resistor. Heat flows only from a hotter body to a colder one, and the water is colder than the resistor. The energy the resistor receives comes from the source as electrical work, not as heat.
• (b) Yes. The resistor is hotter than the water, so heat flows from the resistor into the water.
• (c) Yes. The battery (source) does electrical work on the resistor, \(W = VIt\) (or \(I^2Rt\)), in pushing charge through it. In the sign convention \(Q = \Delta U + W\), this work is done on the system, so \(W\) is negative.
• (d) The state of the resistor does not change, so \(\Delta U = 0\). The first law \(Q = \Delta U + W\) then gives \(Q = W\): both are negative, meaning the system gives out as heat (to the water) exactly the energy it receives as electrical work.