Question

100. mL of water at 40.0 degrees C are added to a calorimeter containing 100. mL water at 20.0 degrees C. After mixing, the water temperature is 29.5 degrees C. Calculate Kc.

Answer 1

Since density of water = 1g/mL,

mass of 100 mL of water, m = 100 mL x1g/mL = 100 g

specific heat capacity of water, s = 4.184 J/C.g

Here water at 40.0 DegC is hotter than water at 20.0xDegC.

Hence heat is released by water at 40 DegC and heat is absorbed by water at 20.0 DegC.

Heat released by water at 40 DecgC = mxsxdT = 100 g x4.184 J/C.g x (40.0 - 29.5) = 4393.2 J

Heat absorbed by water at 20 DecgC = mxsxdT = 100 g x4.184 J/C.g x (29.5 - 20) = 3974.8 J

Since the heat absorbed is less than the heat released, some of the heat is absorbed by the calorimeter.

How ever here there is no equilibrium is maintained between any phase and also there is no chemical reaction in equilibrium. Hence there would be no equilibrium constant Kc