Answer:
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
[tex]H=i^2\times R\times t[/tex]
Where, i = current
R = resistor
t = time
Put the value into the formula
[tex]H=0.5^2\times2\times100[/tex]
[tex]H=50\ J[/tex]
We need to calculate the change in temperature
Using formula of temperature
[tex]\Delta T=\dfrac{50}{0.24}[/tex]
[tex]\Delta T=208.33\ K[/tex]
Now, The temperature is
[tex]T_{2}=300+208.33=508.3\ K[/tex]
We need to calculate the entropy change of the system
Using formula of entropy
[tex]\Delta S=\int_{T_{1}}^{T_{2}}{C\dfrac{dT}{T}}[/tex]
[tex]\Delta S=C\times ln(\dfrac{T_{2}}{T_{1}})[/tex]
[tex]\Delta S=0.24\times ln{\dfrac{508.3}{300}}[/tex]
[tex]\Delta S=0.1265\ J/K[/tex]
Hence, The entropy change of the system is 0.1265 J/K.
