Question #806
How many calories of heat are released as 200.0g of water cool from a liquid at 25.0C to ice at -15.0C?
Expert's answer
According to the heat equations:
Q = C1mΔT1 + C2mΔT2 + λm.
m = 0.2 kg
For water:
C1 = 4.183x103 [J / (kg K)]
ΔT1 = 25 - 0 = 25
For ice:
C2 = 2.11x103 [J / (kg K)]
ΔT2 = 0 - (-15) = 15
λ = 3.3x105 J /kg
Thus
Q = 0.2 (4.183x103*25 +3.3x105 + 2.11x103 *15) = 93.245 J
In calories: Q = 22.29 kcal.
Q = C1mΔT1 + C2mΔT2 + λm.
m = 0.2 kg
For water:
C1 = 4.183x103 [J / (kg K)]
ΔT1 = 25 - 0 = 25
For ice:
C2 = 2.11x103 [J / (kg K)]
ΔT2 = 0 - (-15) = 15
λ = 3.3x105 J /kg
Thus
Q = 0.2 (4.183x103*25 +3.3x105 + 2.11x103 *15) = 93.245 J
In calories: Q = 22.29 kcal.
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#340153 on Dec 2023
#340153 on Dec 2023