Calculate the Calorimeter Constant (Ccal) if 25.00 g of water at 51.50°C was added to 25.00 g of water at 25.00 °C with a resulting temperature of 35.00 °C?

Question

(*remember the specific heat capacity of water (cwater) = 4.180 J/g°C)

Answer 1

Hot water lost
Δ H₁=CmΔT₁ =4.180• 25•(51.5-35) =1724 J
Cold water got
Δ H₂=CmΔT₂ =4.180• 25•( 35-25) =1045 J
Δ H = Δ H₁-Δ H₂=1724-1045=679 J
heat capacity of the calorimeter (Calorimeter Constant)
C(cal)=Δ H/ΔT = 679/10 =67.9 J/℃