Question

Calculate the calorimeter constant for trial 1 and trial 2.

Trial 1:

Hot water: 46.6 C & 50.02 mL

Cold Water: 22.7 C & 50.03 mL

Mixed: 35.3 C

Trial 2:

Hot Water: 55.4 C & 50.00 mL

Cold Water: 24.1 C & 49.09 mL

Mixed: 40.2 C

Answer 1

Trial 1

1) Energy lost by the hot water:

q = m C_p ΔT

Q = ? m = 50.02 gm C_p = 4.184 J/g^-1 °C^-1   ΔT = 46.6 °C - 35.3 °C =  11.3 °C

q = (50.02 g) (4.184 J/g^-1 °C^-1) (11.3 °C)

q = 2364.90 J

2) Energy gained by the cold water:

Q = ? m = 50.03 gm C_p = 4.184 J/g^-1 °C^-1   ΔT = 35.3 °C - 22.7 °C =  11.3 °C

q = m C_p ΔT

q = (50.03 g) (4.184 J/g^-1 °C^-1) (12.6 °C)

q = 2637.50 J

From the above calculation, energy gained by cold water is more than energy lost by the hot water and it is not possible. Hence some calculation error was occurred.

Trial 2

1) Energy lost by the hot water:

q = m C_p ΔT

Q = ? m = 50.00 gm C_p = 4.184 J/g^-1 °C^-1   ΔT = 55.4 °C - 40.2 °C =  15.2 °C

q = (50.00 g) (4.184 J/g^-1 °C^-1) (15.2 °C)

q = 3179.84 J

2) Energy gained by the cold water:

Q = ? m = 49.09 gm C_p = 4.184 J/g^-1 °C^-1   ΔT = 40.2 °C - 24.1 °C =  16.1 °C​​​​​​​

q = m C_p ΔT

q = (49.09 g) (4.184 J/g^-1 °C^-1) (11.2 °C)

q = 3306.82 J

From the above calculation, energy gained by cold water is more than energy lost by the hot water and it is not possible. Hence some calculation error was occurred.