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A steady device in each cycle converts 155 J of work to thermal energy that is released into a reservoir at 340 K. Part A Calculate the change in entropy of the environment that results from this transfer.

Respuesta :

Answer:

change in entropy is 3.3034 × [tex]10^{22}[/tex]  

Explanation:

give data

thermal energy Q  = 155 J

temperature T = 340 K

to find out

change in entropy

solution

we know change in entropy formula that is

change in entropy = Q / ( K×T )   ..............1

here K is boltzmann constant that is 1.38 × [tex]10^{-23}[/tex] kg-m²/s²

put these value in equation 1 we get

change in entropy = Q / ( K × T )

change in entropy = 155 / (  1.38 × [tex]10^{-23}[/tex]  × 340 )

change in entropy = 3.3034 × [tex]10^{22}[/tex]  

so change in entropy is 3.3034 × [tex]10^{22}[/tex]  

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