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?

Answer 1

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

M*C*ln(T2/T1)

where T1 and T2 are initial and final absolute temperatures.

Answer 2

Thank you- this is very helpful