A 100-g aluminum calorimeter contains 500 g of water at a temperature of 20ºC.?
2019-03-08 11:54 am
A 100-g aluminum calorimeter contains 500 g of water at a temperature of 20ºC. Then a 140-g piece of metal, initially at 344ºC, is added to the calorimeter. The metal, water, and calorimeter reach an equilibrium temperature of 32ºC. What is the specific heat capacity of the metal? The specific heat capacity of aluminum is 910 J/kg ∙ K, and of water is 4190 J/kg ∙ K.
回答 (3)
2019-03-08 1:50 pm
Heat change = m c ΔT
Heat lost by the metal = Heat gained by aluminum calorimeter + Heat gained by water
(140)g × c × (344 - 32)°C = (100 g) × (0.910 J/g°C) × (32 - 20)°C + (500 g) × (4.190 J/g°C) × (32 - 20)°C
Specfic heat capacity of the metal, c = (100 × 0.910 × 12 + 500 × 4.19 × 12) / (140 × 312) J/g°C = 0.601 J/g°C = 601 J/kg°C
Heat lost by the metal = Heat gained by aluminum calorimeter + Heat gained by water
(140)g × c × (344 - 32)°C = (100 g) × (0.910 J/g°C) × (32 - 20)°C + (500 g) × (4.190 J/g°C) × (32 - 20)°C
Specfic heat capacity of the metal, c = (100 × 0.910 × 12 + 500 × 4.19 × 12) / (140 × 312) J/g°C = 0.601 J/g°C = 601 J/kg°C
2019-03-08 1:50 pm
(910 J/kg ∙ K) x (0.100 kg Al) x (32 - 20)ºC = 1092 J gained by the Al
(4190 J/kg ∙ K) x (0.500 kg H2O) x (32 - 20)ºC = 25140 J gained by the water
1092 J + 25140 J = 26232 J total gained by the whole calorimeter (and lost by the metal)
(26232 J) / (0.140 kg) / (344 - 32)ºC = 601 J/kg ∙ K
(4190 J/kg ∙ K) x (0.500 kg H2O) x (32 - 20)ºC = 25140 J gained by the water
1092 J + 25140 J = 26232 J total gained by the whole calorimeter (and lost by the metal)
(26232 J) / (0.140 kg) / (344 - 32)ºC = 601 J/kg ∙ K
2019-03-08 12:11 pm
Hint: conservation of energy
收錄日期: 2021-05-01 22:28:57
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