Question

What is the change in entropy for a system that has 250 J of heat added to it at 25°C?

-10 J/K, -0.84 J/K, 0.84 J/K, or 10 J/K

**thank you!!

Answer 1

it is simple, we have to apply this formula:
\[\Large \Delta S = \frac{Q}{T}\]
where \delta S is the entropy change

Answer 2

okay!! so 250/25 = 10 ? so our solution is 10 J/K ?

Answer 3

no, since T is the absolute temperature

Answer 4

ohhh i am unsure then :/

Answer 5

we have:
\[T = t + 273\]
t is the Celsius temperature

Answer 6

ohh so we would get 273+25=298?

Answer 7

that's right!

Answer 8

250/298=0.84?

so our solution is 0.84 J/K?

Answer 9

correct!

Answer 10

yay! thank you!! :D

Answer 11

thank you!! :D