Question

I need some help 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.

Answer 1

12.3°C

Step-by-step 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