# Answer to Question #806 in Physical Chemistry for tristan

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

m = 0.2 kg

For water:

C

ΔT

For ice:

C

ΔT

λ = 3.3x10

Thus

Q = 0.2 (4.183x10

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 = 25For ice:

C

_{2}= 2.11x10^{3}[J / (kg K)]ΔT

_{2}= 0 - (-15) = 15λ = 3.3x10

^{5}J /kgThus

Q = 0.2 (4.183x10

^{3}*25 +3.3x10^{5}+ 2.11x10^{3}*15) = 93.245 J**In calories: Q = 22.29 kcal.**Need a fast expert's response?

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