Calculate the calorimeter constant in the following case:A student adds 50.0 ml H2O at 48.5 degrees Celsius to 50.0 ml H2O at 23.2 degrees Celsius in a calorimeter. The temperature of the...

Calculate the calorimeter constant in the following case:

A student adds 50.0 ml H2O at 48.5 degrees Celsius to 50.0 ml H2O at 23.2 degrees Celsius in a calorimeter. The temperature of the mixture was 35.4 degrees Celsius.  (Density of H2O is 1.00 g\ml and specific heat of H2O is 4.18 J\g*C)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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To determine the calorimeter constant, a student adds 50.0 ml H2O at 48.5 C to 50.0 ml H2O at 23.2 C in a calorimeter. The temperature of the mixture that is obtained is 35.4 C. The density of water is 1g/ml and its specific heat is 4.18 J/g*C

When the two samples of water each with a volume of 50 mL, one at 48.5 C and the other at 23.2 C are mixed, the temperature of the resulting mixture should be (48.5+23.2)/2 = 35.85 C. Instead it is 35.4 C. The amount of heat taken up by the calorimeter from 50 mL of the heated water added is 4.18*0.45*100 = 188.1 J

The heat required to change the temperature from 23.2 to 35.4 C is 188.1 J. This gives the calorimeter constant as 15.41 J/C

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