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If an aluminum cube of mass 47.64 g is heated to 100.0°C and is then immersed...

If an aluminum cube of mass 47.64 g is heated to 100.0°C and is then immersed in water and allowed to equilibrate to a final temperature of 32.4°C then how much heat did the aluminum cube lose? (Q < 0) Assume 0.215 cal/g°C as the specific heat of aluminum. Report your answer to two (2) decimal places.

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Answer #1

Spe = 0.215 cal 1 goc Q=MSAAT - ore 47.644 | 32.4 t0) x 0.215 =-692.399 Call -692.40 cal = 692:40 x 4:18 4 J = –2899 J

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