How much heat (kJ) is absorbed by 193.7 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?
回答 (1)
2017-02-07 2:13 pm
Mass of water, m = 193.7 g
Specific heat capacity of water, c = 4.186 J/g°C
Temperature change of water, ΔT = (32.50 - 25.00)°C = 7.50°C
Heat absorbed = m c ΔT = (193.7 g) × (4.186 J/g°C) × (7.50°C) = 6080 J = 6.08 kJ
(3 sig. fig.)
Specific heat capacity of water, c = 4.186 J/g°C
Temperature change of water, ΔT = (32.50 - 25.00)°C = 7.50°C
Heat absorbed = m c ΔT = (193.7 g) × (4.186 J/g°C) × (7.50°C) = 6080 J = 6.08 kJ
(3 sig. fig.)
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