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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

Sagot :

whitneytr12

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

[tex]\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})[/tex]

Where:

Q: is the energy transferred = 5.0 MJ

[tex]k_{B}[/tex]: is the Boltzmann constant = 1.38x10⁻²³ J/K  

[tex]T_{i}[/tex]: is the initial temperature = 1000 K

[tex]T_{f}[/tex]: is the final temperature = 500 K

Hence, the entropy change is:

[tex] \Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26} [/tex]                                    

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

I hope it helps you!          

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