posted by prabhu on .
A thermally insulated 50 ohm resistor carries a current of 1 A for 1 s. The initial temperature of the resistor is 10 degree Celsius, its mass is 5 g, its specific heat capacity is 850. What is the change in entropy of the resistor?
First calculate how much the temperature rises. Neglect heat loss during the brief interval of resistive heating. The deposited heat energy is
Q = I^2*R*t = 50 Joules.
You have not said what your units of heat capacity (C) are, so I cannot tell you what the temperature rise is. The formula to use is:
(T2 - T1) = Q/(M*C)
The entropy change is the integral of dQ/T, which is
where T1 and T2 are initial and final absolute temperatures.