A 92.5 g piece of aluminum (which has a molar heat capacity of 24.03]/°C-mol) is heated...

Question

Question

Answer 1

Answer #1

Given data

Mass of metal Al = 92.5 g

Molecular weight of Al = 26.98g/mol

specific heat capacity of metal = 24.03 J/°C-mol

Initial temperature of Al = 624 °C

Final temperature of Al = 135.2 °C

specific heat capacity of water = 4.18 J/g-°C

Initial temperature of water = 19.2 °C

Final temperature of water = 135.2 °C

Heat released by metal = heat absorbed by water

Mass of metal x specific heat capacity of metal x temperature change = Mass of water x specific heat capacity of water x temperature change

(92.5 g / 26.98g/mol) x 24.03 J/°C-mol x (624 - 135.2)°C = Mass of water x 4.18 J/g-°C x (135.2 - 19.2)°C

3.428 mol x 24.03 J/°C-mol x 488.8°C = Mass of water x 4.18 J/g-°C x 116 °C

Mass of water = 40264.821 / 519.68

= 77.5 g

Answer 2

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