Answer to Question #806 in Physical Chemistry for tristan

Question

How many calories of heat are released as 200.0g of water cool from a liquid at 25.0C to ice at -15.0C?

Answer 1

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.