Posted by prabhu on Wednesday, February 1, 2012 at 12:10am.
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?

science  drwls, Wednesday, February 1, 2012 at 2:30am
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.
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