Question

54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the system is 59.4 oC, what is the calorimeter constant (C calorimeter)

Answer 1

Answer:

49.5J/°C

Explanation:

The hot water lost some energy that is gained for cold water and the calorimeter.

The equation is:

Q(Hot water) = Q(Cold water) + Q(Calorimeter)

Where:

Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J

Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J

That means the heat gained by the calorimeter is

Q(Calorimeter) = 4794J - 3834J = 960J

The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:

59.4°C-40°C = 19.4°C

And calorimeter constant is:

960J/19.4°C =

49.5J/°C