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

Question

A student determined the calorimeter constant of the calorimeter, using the procdure 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 19.20 °C and of the hot water, 29.60 °C. The maximum temperature of the mixture was found to be 24.75 °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 T of the cold water.

Explanation / Answer

heat lost by hot water = 50 x 4.184 x (24.75 - 19.20) = 1161.06 J

heat gained by cold water = 50 x 4.184 x (29.60 - 24.75) = 1014.62 J

Heat absorbed by calorimeter = 1161.06 - 1014.62 = 146.44 J

Calorimeter constant = 146.44/(24.75 - 19.20) = 26.385 J/oC

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