## anonymous one year ago 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!!

1. Michele_Laino

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

2. anonymous

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

3. Michele_Laino

no, since T is the absolute temperature

4. anonymous

ohhh i am unsure then :/

5. Michele_Laino

we have: $T = t + 273$ t is the Celsius temperature

6. anonymous

ohh so we would get 273+25=298?

7. Michele_Laino

that's right!

8. anonymous

250/298=0.84? so our solution is 0.84 J/K?

9. Michele_Laino

correct!

10. anonymous

yay! thank you!! :D

11. Michele_Laino

thank you!! :D