n the calibration step of a thermochemistry experiment, a current of 124/text{ mA}124 mA, from a 24.0/text{ V}24.0 V source was allowed to flow through the electrical heater for 219/text{ s}219 s and was found to result in an increase in the temperature of the calorimeter and its contents of + 1.39/text{ K}+1.39 K. Given that the energy supplied by an electrical circuit is the voltage multiplied by the current, multiplied by the time, what is the heat capacity of the calorimeter and its contents?

Question

Anya Robbins

Physics

24 September, 03:22

n the calibration step of a thermochemistry experiment, a current of 124/text{ mA}124 mA, from a 24.0/text{ V}24.0 V source was allowed to flow through the electrical heater for 219/text{ s}219 s and was found to result in an increase in the temperature of the calorimeter and its contents of + 1.39/text{ K}+1.39 K. Given that the energy supplied by an electrical circuit is the voltage multiplied by the current, multiplied by the time, what is the heat capacity of the calorimeter and its contents?

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Zechariah Hickman · Answer 1

Answer: C = 468.88 J/kgk

Explanation: Given that the

Current I = 124 mA

Voltage V = 24 V

Time t = 219 s

Change in temperature Ø = 1.39 K

Heat energy = mcØ

Also,

Heat energy = IVt

Where c = specific heat capacity

IVt = mcØ

Let assume that the mass is negligible or a unit mass = 1kg

124 * 10^-3 * 24 * 219 = 1.39c

C = 651.744/1.39

C = 468.88 J/kgk