) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a 24.0 V source was allowed to flow through the electrical heater for 247 s and was found to result in an increase in the temperature of the calorimeter and its contents of +1.25 K. Calculate the heat capacity of the calorimeter and its contents

Explanation:

Heat energy supplied = IVt where I = current = 117 A / 1000 = 0.117 A

V = 24 V and t, time = 247 s

Heat energy supplied = 0.117 A × 24V × 247s = 693.576 J

Heat capacity = quantity of heat supplied / rise in temperature

Heat capacity = 693.576 J / 1.25 K = 554.86 J/K

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