March 30, 2017

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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?

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

  • Please help how do I calculate this - ,

    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/℃

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