Calculate the amount of heat required to raise the temperature of a 48g sample of water from 53°C to 88°C.?
回答 (3)
2017-09-12 7:10 pm
Amount of heat required, E
= m c ΔT
= (48 g) × (4.184 J/g°C) × [(88 - 53)°C]
= 7000 J
= 7.0 kJ
= m c ΔT
= (48 g) × (4.184 J/g°C) × [(88 - 53)°C]
= 7000 J
= 7.0 kJ
2017-09-12 7:33 pm
(48 g)(1 cal/gK)(35 K) = 1680 calories or 1.68 kcal
2017-09-12 7:23 pm
Use the equation
Q = mcDeltaT
Where
m = 48g
C = 4.18 (Specific Heat of water)
DeltaT = 88 - 53 = 35
Hence
Q = 48 X 4.18 x 35 = 7022.4 J = (+)7.0224 kJ
Q = mcDeltaT
Where
m = 48g
C = 4.18 (Specific Heat of water)
DeltaT = 88 - 53 = 35
Hence
Q = 48 X 4.18 x 35 = 7022.4 J = (+)7.0224 kJ
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