A 58.36 g sample of water at 71.4 degrees C is added to a sample of water at 23.2 degrees C in a constant-pressure calorimeter. If the final temperature of the combined water is 41.9 degrees C and the heat capacity of the calorimeter is 26.3 J/C, calculate the mass of the water originally in the calorimeter.

There are two heat transfers happening according to this question. The heat transfer from external water sample to the calorimeter and from external water sample to calorimeter water sample.

Now, as the equilibrium will be achieved, the heat transfer of external water sample will be equal to the sum of above mentioned heat transfers but in opposite direction. Such as:

Qext.w = - (Qcal + Qw.cal) ------- (1)

Since, Q = mCp ΔT

Where,

Q = Heat

m = Mass

ΔT = Change in Temperature

1. Qext.w = (mext.w) (Cpext.w) ΔText.w = Heat transfer from ext. water

2. Qcal = (Cpcal)(ΔTcal) = Heat transfer to calorimeter

3. Qw.cal = (mw.cal) (Cpw.cal) ΔTw.cal = Heat transfer to water already in calorimeter

4. Cp of water = 4.12 J/g.°C

Substituting the values in equation (1), we get,

(58.36g)(4.12J/g. °C)(41.9-71.4) = - (26.3 J/°C) (41.9-23.2)

- mw.cal (4.12 J/g.°C)(41.9-23.2)

mw.cal (4.12 J/°C) (41.9-23.2) = - (26.3 J/°C) (41.9-23.2)

- (58.36g)(4.12J/g. °C)(41.9-71.4)

mw.cal (4.12) (41.9-23.2) = -491.81 - (-7093.07)

mw.cal (4.12) (41.9-23.2) = 6601.26

mw.cal = 6601.26 / [(4.12) (41.9-23.2)]

mw.cal = 85.68 g