Acurrent of 0,5a flows through a 22 resistor for 100s. the system who- se initial temperature is 300k is thermally isolated. the heat capacity of the resistor is 0,24jk-1 for a wide range of temperatures. the temperatu- re of the resistor changes appreciably. what is the entropy change of the system?

The entropy change of the system is 0.1265 J/K.

Explanation:

Given that,

Current = 0.5 A

Resistor = 2 Ω

Time = 100 s

Temperature = 300 K

Heat capacity = 0.24 J/K

We need to calculate the heat produced

Using formula of heat

Where, i = current

R = resistor

t = time

Put the value into the formula

We need to calculate the change in temperature

Using formula of temperature

Now, The temperature is

We need to calculate the entropy change of the system

Using formula of entropy

Hence, The entropy change of the system is 0.1265 J/K.

nah.   i'm just sleepy.

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