I know the answer is 282 J/degree C but i don't understand why?

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I know the answer is 282 J/degree C but i don't understand why?

A calorimeter is calibrated with an electrical heater. Before the heater is turned on, the
calorimeter temperature is 25.3 0C. The addition of 1.07 kJ of energy from the heater raises the
temperature to 29.1 0C. What is the total heat capacity of the calorimeter?

2007-10-16 14:02:27 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

1 answers

1070J/(29.1C - 25.3C) = (1070/3.8) J/C = 282 J/C

2007-10-19 16:20:15 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋