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Solve heat capacity problems Question An object with a heat capacity of 3.40 x 103. absorbs 54.0 kJ of heat, beginning at -25
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- Absorbed heat= Heat capacity x tempo change 9 suxioʻI = 3.40x/0x x [18-1-25°)] N 7 3 40X103 I/O. Tp +25°C = 15.88°C TA 9.1°

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