Answer to Question #107617 in Physical Chemistry for Nouf

Question

The following information was recorded for a calorimeter, complete the given table,
and calculate e the heat capacity heat of water)
Volume of cold water T1 (cold water) Volume of hot water T2(hot water)
Tf
of the calorimeter (use 4.184 J g-1 °C-1 as the specific
100 ml 23°C 100 ml 48°C 35.5°C
∆t (cold water)
................
∆t (hot water)
................
Q\ (cold water)
...............
Q\ (cold water)
..................

Answer 1

m (water) = 100 g

T1 = 23

T2 = 48

Tf = 35.5

∆t (cold water) = 35.5 - 23 = 12,5 ⁰C

∆t (hot water) = 48 - 35.5 = 12.5 ⁰C

Q (cold water) = cm∆t (cold water) = 12.5⁰C*100 g * 4.184 J /g-1 °C-1 = 5230 J = 5.23 kJ

Q (hot water) = cm∆t (hot water) = 12.50C*100 g * 4.184 J /g-1 °C-1 =5230 J = 5.23 kJ