SSPart E. Determination of a Calorimeter Constant A student doing a calorimetry

Question

SSPart E. Determination of a Calorimeter Constant A student doing a calorimetry experiment similar to this one is asked to find the calorimeter constant for the calorimeter he is using. Cold water is added to the calorimeter, the temperature is measured. Water is heated on a hot plate. The temperature is measured. Some hot water is added to the calorimeter with the cold water. The final temperature is measured. The masses are taken along the way. The data is given below. Trial 1 25.4356 Trial 2 Trial 3 25.6738 50.2011g 19.9°C 90.2°C 75.0680g Mass of empty calorimeter 25.4361g 50.19388 20.1 85.6°C 74.3392g 50.1927g Mass calorimeter with cold water Temperature of cold water in calorimeter 19.7°C Temperature of hot water Mass of calorimeter with hot and cold water mixture Final temperature of water mixture | 83.5 75.2201g 48.9°C 49.5°C 52.2°C Use the data given above to calculate the calorimeter constant including the average and standard deviation. Be sure to give correct units and significant figures. As always, work must be shown to receive credit.

Explanation / Answer

From the given calorimeter data,

Trial 1,

Heat gained by cold water = mCpdT

m = 24.7571 g

Cp = 4.184 J/g.oC

dT = 48.9 - 19.7 = 29.2 oC

So, heat gained = 24.7571 x 4.184 x 29.2 = 3024.64 J

Heat lost by cold water = mCpdT

m = 24.0274 g

Cp = 4.184 J/g.oC

dT = 83.5 - 48.9 = 34.6 oC

So, heat gained = 24.0274 x 4.184 x 34.6 = 3623.13 J

heat gained by calorimeter = 3623.13 - 3024.64 = 598.5 J

calorimeter constant = 598.5 J/29.2 oC = 20.5 J/oC

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Trial 2,

Heat gained by cold water = mCpdT

m = 24.7577 g

Cp = 4.184 J/g.oC

dT = 29.4 oC

So, heat gained = 24.7577 x 4.184 x 29.4 = 3045.43 J

Heat lost by cold water = mCpdT

m = 24.1454 g

Cp = 4.184 J/g.oC

dT = 36.1 oC

So, heat gained = 24.1454 x 4.184 x 36.1 = 3647.0 J

heat gained by calorimeter = 601.6 J

calorimeter constant = 601.6 J/29.4 oC = 20.5 J/oC

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Trial 3,

Heat lost by hot water = mCpdT

m = 24.8669 g

Cp = 4.184 J/g.oC

dT = 90.2 - 52.2 = 38 oC

So, heat lost = 24.8669 x 4.184 x 38 = 3953.64 J

Heat gained by cold water = mCpdT

m = 24.5273 g

Cp = 4.184 J/g.oC

dT = 52.2 - 19.9 = 32.3 oC

So, heat gained = 24.5273 x 4.184 x 32.3 = 3314.70 J

heat gained by calorimeter = 3953.64 - 3314.70 = 638.94 J

calorimeter constant = 638.94 J/32.3 oC = 20.0 J/oC

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mean calorimeter constant = 20.3 J/oC

standard deviation = sq.rt.[(x - mean)^2/3] = sq.rt.[(0.04 + 0.04 + 0.09)/3]

                              = 0.24 J/oC

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