1. Calculate the amount of heat required to heat a 3.0 kg gold bar from 21 ∘C to 62 ∘C. Specific heat capacity of gold is 0.128 J/g∘C.
2. If 60 J of heat is added to an aluminum can with a mass of 26.9 g , what is its temperature change? Specific heat capacity of aluminum is 0.903 J/g∘C.
Chemistry Easy 10 points ?
2015-09-10 9:13 am
回答 (1)
2015-09-10 9:38 am
✔ 最佳答案
1.Heating gold bar :
m = 3.0 kg = 3000 g
c = 0.128 J/g°C
ΔT = (62 - 21)°C = 41 °C
Heat required
= m c ΔT
= (3000 g) × (0.128 J/g°C) × (41°C)
= 15700 J ...... (Ans)
2.
Heating aluminum can :
E = 60 J
m = 26.9 g
c = 0.903 J/g°C
E = m c ΔT
Temperature change, ΔT
= E / (m c)
= (60 J) / [(26.9 g) × (0.903 J/g°C)]
= 2.47°C ...... (Ans)
收錄日期: 2021-05-01 09:38:34
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