The following data was collected for determining the heat capacity of a calorime

Question

The following data was collected for determining the heat capacity of a calorimeter. The temperature of 150 ml. of cool water in the calorimeter was measured as 21, 5 degree C. The temperature of 100 ml. of warm water was 35.5 degree C. The temperature of the mixture extrapolated to the time of mixing was 26.8 degree C. Calculate the heat capacity of the calorimeter. A 173 g lead ball was heated to 100 degree C and plunged into 250 ml. of 19 degree C water. The specific heat of lead is 0.1276 J/g degree C. the specific heat of water is 4.184 J/g degree C. Calculate the final temperature of the water. Ignore the heat capacity of the calorimeter. In the heat of neutralization experiment, a student could only obtain a thermometer 7 readable to plusminus 0.5 degree C. Calculate the maximum expected error in percent (relative I precision as %) which could result than this piece of equipment, assuming the error is due only to the readability of the thermometer. Use the data in the calculation example (see Sources of Error rightarrow Equipment Error rightarrow The thermometer.) A 47.8 g piece of metal is heated to 104.3 degree C and plunged into an insulated vessel containing 65.4 g water at 26.5 degree C. If the container has a heat capacity of 8.7 J/K. and the final temperature both the water and container is 34.2 degree C. what is the specific heat of the metal?

Explanation / Answer

1. heat absorbed by cool water = 150 x 4.184 x (26.8 - 21.5)

= 3326.28 J

heat released by warm water = 100 x 4.184 x (35.5 - 26.8)

= 3640.08

heat absorbed bt calorimeter = 3640.08 - 3326.28 = 313.8 J

313.8 = C (26.8 - 20)

C = 46.2 J / deg C .....Ans

2. 173 x 0.1276 x (100 - T) = 250 x 4.184 x (T - 19)

2207.5 - 22.07 T = 1046T - 19874

T = 20.7 deg C

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