Question

A current 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?

Answer 1

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

$H=i^2\times R\times t$

Where, i = current

R = resistor

t = time

Put the value into the formula

$H=0.5^2\times2\times100$

$H=50\ J$

We need to calculate the change in temperature

Using formula of temperature

$\Delta T=\dfrac{50}{0.24}$

$\Delta T=208.33\ K$

Now, The temperature is

$T_{2}=300+208.33=508.3\ K$

We need to calculate the entropy change of the system

Using formula of entropy

$\Delta S=\int_{T_{1}}^{T_{2}}{C\dfrac{dT}{T}}$

$\Delta S=C\times ln(\dfrac{T_{2}}{T_{1}})$

$\Delta S=0.24\times ln{\dfrac{508.3}{300}}$

$\Delta S=0.1265\ J/K$

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