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Suppose you negleted to account for heat transferred to the caliorimeter. What percentage error would be...

Suppose you negleted to account for heat transferred to the caliorimeter. What percentage error would be intorduced to your measurement of the specific heat of copper?
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Answer #1

If you neglect the heat transferred to the calorimeter, the measure of specific heat will be given by the equatiom:

Specific heat = Heat transferred to water / (mass of Copper * temeprature change)

Since we have neglected the heat transferred to calorimeter, the numerator is lower than the actual Heat released meaning that the specific heat will be lower

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