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 = C_{1}mΔT_{1} + C_{2}mΔT_{2} + λm.

m = 0.2 kg

For water:

C_{1} = 4.183x10^{3} [J / (kg K)]

ΔT_{1} = 25 - 0 = 25

For ice:

C_{2} = 2.11x10^{3} [J / (kg K)]

ΔT_{2} = 0 - (-15) = 15

λ = 3.3x10^{5} J /kg

Thus

Q = 0.2 (4.183x10^{3}*25 +3.3x10^{5} + 2.11x10^{3} *15) = 93.245 J

**In calories: Q = 22.29 kcal.**

Q = C

m = 0.2 kg

For water:

C

ΔT

For ice:

C

ΔT

λ = 3.3x10

Thus

Q = 0.2 (4.183x10

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