A 150.0-g sample of metal at 80.0ºC is added to 150.0 g of H2O at 20.0ºC. The temperature rises to23.3ºC. Assuming that the calorimeter is a perfect insulator, what is the specific heat of the metal? (Specific heat of H2O is 4.18 J/g · ºC.)
A)–0.48 J/g · ºC
B)0.24 J/g · ºC
C)0.48 J/g · ºC
D)0.72 J/g · ºC
E)0.96 J/g · ºC
I need help for this one. explain if you could thank you?
2017-12-10 5:23 pm
回答 (1)
2017-12-10 8:30 pm
✔ 最佳答案
Heat lost from the metal = (150.0 g) × [y J/(g °C)] × [(80.0 - 23.3)°C]Heat gained by the water = (150.0 g) × [4.18 J/(g °C)] × [(23.3 - 20.0)°C]
Heat lost from the metal = Heat gained by the water
150.0 g × y × (80.0 - 23.3) = 150.0 × 4.18 × (23.3 - 20.0)
y = 4.18 × [(23.3 - 20.0)/(80 - 23.3)]
y = 0.243 ≈ 0.24 (to 2 sig. fig.)
Specific heat of the metal = 0.24 J/(g °C)
The answer: B) 0.24 J/(g °C)
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