Thursday
March 30, 2017

Post a New Question

Posted by on .

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/℃

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question