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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?

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    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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