Question

An electric current of 1 A flows for 10 s in a reservoir of resistance 25 W which is submerged in a large volume of water, the temperature of which is 280 K. Calculate (a) the change in the entropy of the resistance, and (b) the change in the entropy of the water.

Answer 1

Because the volume of water is assumed to be large, therefore the temperature of the water does not change. Consequently, the change in heat released by the resistor is zero [Since the current through the resistor is constant].

therefore, the change in entropy of the resistor will be:

Delta S = Delta Q/T = 0 J/K

b]

resistance = 25 ohms [it is 25 ohms and not 25 watts as mentioned in the question]

current = 1 A

time t = 10s

so, the heat liberated to the water is:

Delta Q = i²Rt = (1)²(25)(10) = 250 J

T = 280 K

therefore, change in entropy of the water = Delta S = 250/280 = 0.8928 J/K.