When 47.8 J of heat are added to 13.0 g of a liquid, its temperature rises by 1.77 ∘C. What is the heat capacity of the liquid?
回答 (3)
2017-02-12 2:44 pm
✔ 最佳答案
Energy= mass*c*change in tempC = 47.8/(13*1.77)
C = 2.08
2017-02-12 8:59 pm
Heat added, E = 47.8 J
Mass, m = 13.0 g
Temperature rise, ΔT = 1.77°C
E = m c ΔT
c = E / (m ΔT)
Heat capacity, c = (4.78 J) / [(13.0 g) × (1.77°C)] = 0.208 J / (g °C)
Mass, m = 13.0 g
Temperature rise, ΔT = 1.77°C
E = m c ΔT
c = E / (m ΔT)
Heat capacity, c = (4.78 J) / [(13.0 g) × (1.77°C)] = 0.208 J / (g °C)
2017-02-12 3:25 pm
One calorie is the amount of energy required to raise the temperature of 1 gram of water by 1 degree celsius. One calorie is equal to 4.181 joules.
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