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