You have a 5.82 calorimeter containing some water at roomtemperature (22.0degC).
ID: 688628 • Letter: Y
Question
You have a 5.82 calorimeter containing some water at roomtemperature (22.0degC). When you add 40.33g boiling water, thetemperature raises to 50.9degC. The calorimeter constant is160.8J/degC. Atmospheric pressure is 760mmhg. The specific heat ofwater is 4.184J/gdegC. Show all work,include all unites. Make suresigns are correct. 1.Calculate the heat flow for the hot water. 1b. Calculate the heat flow of the cold water. 1c. Calculate how much cold water was added. You have a 5.82 calorimeter containing some water at roomtemperature (22.0degC). When you add 40.33g boiling water, thetemperature raises to 50.9degC. The calorimeter constant is160.8J/degC. Atmospheric pressure is 760mmhg. The specific heat ofwater is 4.184J/gdegC. Show all work,include all unites. Make suresigns are correct. 1.Calculate the heat flow for the hot water. 1b. Calculate the heat flow of the cold water. 1c. Calculate how much cold water was added.Explanation / Answer
Given that the calorimerter constant = 160.8 J/degC Therefore the heat absorbed by the calorimeter =qcal = CT =160.8 J/deg C (50.9 - 22) deg C = 4647.12 J Let us assume that theheat releasing during the addition of boilingwater q = 40.33 g * 4.184 (50.9-22) = - 4876 J The same amount of heat gained by the calorimeter +water = 4876 J Therefore heat absorbed by the cold water = 4876 J -4647.12 = 228.88 J Hence - 228.88 J = m * 4.184 *(50.9 -22) Thusmass of the cold water = 1.892g Therefore the heat flow of the hot water = 4876 J Heat flow of the cold water = 228.88 J Mass of the cold water in the container = 1.892gRelated Questions
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