Question

a calorimeter initially contains 75.00 g of water at 25.00 degrees celsius. after a 45.00 g sample of nickel, initially at 100.00 degrees celsius, the final temperature of the system was 29.98 degrees celsius. what is the heat capacity of the calorimeter in j/c?

Answer 1

The heat capacity (in J/°C) of the calorimeter, given that the final temperature of the system was 29.98 degree celsius, is 32.88 J/ºC

How to calculate the heat capacity?

First, we shall obtain the heat liberated by the nickel. Deatils below:

• Mass of nickel (M) = 45.00 g
• Initial temperature of nickel (T₁) = 100.00 degree celsius
• Final temperature of nickel (T₂) = 29.98 degree celsius
• Change in temperature of nickel (ΔT) = 29.98 - 100 = -70.02 degree celsius
• Specific heat capacity of nickel (C) = 0.444 J/gºC
• Heat liberated by nickel (Q) =?

Q = MCΔT

Q = 45 × 0.444 × -70.02

Q = -1398.9996 J

Next, we shall calculate the heat absorbed by the water. This is shown below:

• Mass of water (M) = 75.00 g
• Initial temperature of water (T₁) = 25.00 degree celsius
• Final temperature of water (T₂) = 29.98 degree celsius
• Change in temperature of water (ΔT) = 29.98 - 25 = 4.98 degree celsius
• Specific heat capacity of water (C) = 4.18 J/gºC
• Heat absorbed by water (H) =?

H = MCΔT

H = 75 × 4.18 × 4.98

H = 1562.724 J

Finally, we shall calculate the heat capacity of the calorimeter. Details below:

• Heat liberated by nickel (Q) = -1398.9996 J
• Heat absorbed by water (H) = 1562.724 J
• Change in temperature of water (ΔT) = 4.98 degree celsius
• Heat capacity of calorimeter =?

`Heat\ capacity = (H\ +\ Q)/(\Delta T) \\\\Heat\ capacity = (1562.724\ -\ 1398.9996)/(4.98)\\\\Heat\ capacity = 32.88\ J/\textdegree C\\\\`