A student determined the calorimeter constant of the calorimeter, using the proc

Question

A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 21.15 degree C and of the hot water, 29.75 degree C. The maximum temperature of the mixture was found to be 24.05 degree C. Assume the density of water is 1.00 g mL^-1 and the specific heat is 4.184 J g^-1 K^-1. Calculate the calorimeter constant, using the AT of the cold water. Your Answer:

Explanation / Answer

(1) Heat lost by warm water:

Q= m x Cp x T

T = (29.5 – 24.05) °C = 5.7°C

Cp =4.184 J/g °C

m = density x volume = 50ml x 1g/ml = 50

q = (50 g) (5.7°C) (4.184 J/g °C) = 1192.44 J

(2) Heat gained by water in the calorimeter:

T = (24.05 – 21.15) °C = 2.9°C

q = (50 g) (2.9 °C) (4.184 J/g °C) = 606.68 J

3) The difference of heat absorbed by the calorimeter:

1192.44 J - 606.68 J = 585.76J

4) Calorimeter constant:

585.76 J / 2.9 °C = 201.98 J/°C

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