What would be the effect on the calibration factor obtained from an experiment if you used 50mL water instead of 100 mL of water to calibrate the calorimeter?

Asked on by fern001

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valentin68 | College Teacher | (Level 3) Associate Educator

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The calibration factor `F` of a calorimeter is by definition to the heat supplied to the calorimeter divided by the temperature change.

To determine the value of ` ` the calorimeter you heat a known mass of water in the calorimeter from an initial temperature to a final temperature with a known heat `Q` provided by an electric heater supplied with a known voltage and current.
 Thus if `C` is the heat capacity of the calorimeter itself:

`Q =C*Delta(T) +m*c_(water)*Delta(T)`

The calibration factor is

`F = Q/(Delta(T)) (=(U*I*t)/(Delta(T))) = C +m*c_(water)`

`F =C +m*c_(water)`

Usually a calorimeter is made of a metal and has small mass. Thus the value of calorimeter heat capacity `C` is about ten times smaller than the heat capacity of water inside it, `m*c_("water")` . Hence the above relation can be approximated well with

`F ~~m*c_("water")`

Now we have the answer: if we use half of initial mass of water to calibrate the calorimeter the calibration factor will be reduced to about half of its initial value.


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