Question

A 41.33g sample of water at 84.6°C is added to a sample of water at 24.1°C in a constant-pressure calorimeter. If the final temperature of the combined water is 40.1°C and the heat capacity of the calorimeter is 26.3 J°C, calculate the mass of the water originally in the calorimeter.

Answer 1

To solve this, you should know that, o Heat lost by hot water = heat gained by cold water theat gained by calorimeten. F 2 hot = a cold + a call amo r e where :-) selgu means that hot water is losing heat. that = (medt) het I cald = (MCDT) cold Ical - CAT where, m = mass of water - speuple heat of water AT = change in Temperature C- Heat capauty of the calorimeter Few Given values: man of not water = 41.339 Let man of cold water : m specife heat of wate= 4:1845|gxi Heat capauty of Calorimeter = 26.3 51°C DT hot, - Final Temp. Pristal Temp. DT hot = 40.16 - 84.6°C = -44.5 DT cold = 40.16 - 24.1°C = 16°E Now Keep values of all these in eqmo.

ghet : Reald + Real - Come st) hot (medt) cold + COT -141.33 g) (4.184 ) (-44.5°c) = m(41845-) ( 16°c) + 26.3 I ( 16°) 7695.155 um 66.945 +420.85 1695.155-420.85 66.94 5 . m = 108.67 g 52 Hence mais of water anginally in the calorimeter is [108.679)