A 25.0 g block of iron at 67.45 °C is added to 100.0 g of water at 25.00 °C. What will be the final temperature of the iron and water at thermal equilibrium? [Hint: the heat given off by the iron must be equal in magnitude but opposite in sign to the heat absorbed by the water.]
Can someone please explain how to solve this?
I got as far as:
-[m * s * delta t] = m * s * delta t
-[(25.0g) * s * (tfinal - 67.45 °C)] = (418 J/°C) (tfinal-25.00 °C)
Additional Details:
This was a quiz question and the specific heat of iron was not given to us so I have to solve this without knowing it. [My chem professor told us we don't need to memorize any specific heats besides water so if it wasn't given that it shouldn't be part of the method used to solve the question.]
Answers & Comments
Verified answer
You cannot solve this without the specific heat of iron.
Call the final temperature x
Heat lost by iron = specific heat*mass*temp drop = S*25*(67.45 - x)
Heat gained by water = specific heat*mass*temp rise = 4.18*100*(x - 25.00)
Heat lost by iron = Heat gained by water
S*25*(67.45 - x) = 4.18*100*(x - 25.00)
S = the specific heat of iron = 0.45 J/gK (and K = C for a temp change)
0.45*25*(67.45 - x) = 4.18*100*(x - 25.00)
758.813 - 11.25x = 418x - 10450
11208.8 = 429.25 x
x = 11208.8//429.25 = 26.11 C