Question

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one at 500 K

Answer 1

The entropy change in the environment is 3.62x10²⁶.

Step-by-step explanation:

The entropy change can be calculated using the following equation:

`\Delta S = (Q)/(k_(B))((1)/(T_(f)) - (1)/(T_(i)))`

Where:

Q: is the energy transferred = 5.0 MJ

`k_(B)`: is the Boltzmann constant = 1.38x10⁻²³ J/K

`T_(i)`: is the initial temperature = 1000 K

`T_(f)`: is the final temperature = 500 K

Hence, the entropy change is:

`\Delta S = (5.0 \cdot 10^(6) J)/(1.38 \cdot 10^(-23) J/K)((1)/(500 K) - (1)/(1000 K)) = 3.62 \cdot 10^(26)`

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!