- we provided box. A 31.5-g sample of an alloy at 93.00°C is placed into 50.0...

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Answer 1

Answer #1

heat lost by the alloy = heat gained by the water & cup

m C dT alloy = m C dT water + C dT cup

(31.5 g )(C)(93.00- 31.00) = (50.0 g)(4.184 J/g-C)(31.10 – 22.00) + (9.20 J/C)(31.10 – 22.00)

(31.5 g )(C)(61.9 C) = (1903.72 ) + (83.72) J

1949.85 g C = 1987.44 J

C= 1.019 J/ g C

Answer 2

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