Ineed some with this: a 0.225 kg block of iron at -28.7 degrees celsius is put in a cup of 0.150 kg of water at 18.9 degrees celsius. what is their equilibrium temperature?
i got the answer 28.1 degrees celsius, but i'm pretty sure that is incorrect.

12.3°C

Explanation:

Logically, if the iron is initially at -28.7°C, and the water is at 18.9°C, then the equilibrium temperature must be somewhere in between.  We can find that temperature using an energy balance.

Heat lost by water = heat gained by iron

-mCΔT = mCΔT

-(0.150 kg) (4.184 kJ/kg/K) (T − 18.9°C) = (0.225 kg) (0.450 kJ/kg/K) (T − (-28.7°C))

(0.628 kJ/K) (18.9°C − T) = (0.101 kJ/K) (T + 28.7°C)

11.9 kJ − (0.628 kJ/K) T = (0.101 kJ/K) T + 2.91 kJ

8.96 kJ = (0.729 kJ/K) T

T = 12.3°C

ill do it for 100, also this isnt 80 its 40

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: