Question

Chapter 12, Question 11 Ice melts at 273.15 K with ΔHfus = 6.01 kJ/mol. An ice cube of mass 34.4 g is dropped into a swimming pool, the temperature of which is held at 27.2°C.

(a) What is the entropy change (ΔS) for melting the ice?

ΔS = ------J/K

(b) What is ΔS of the pool?

ΔS = ------J/K

(c) What is the overall ΔS?

ΔS = ------J/K

Answer 1

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- When a system received an amount of q energy at a const Temperature, The eorthalpy change (as) defined as - asan. Alfus , n = mole no. - T. K Alfus. = enthalpy of . fusion hecਹਾ , I e Temperature. @ 4s (entropy change for melting of ice): amount of ice = 34.4 9. E- 3 4.4 18 .:. Amont of icebna 1911 mol. T=273:15 x ; Allfws = 6.01% 16 /mol as cl 1.911 X 6010 61273.15-) J/K. = 42.049 J/K. © For the melting of ice, the supplied heat is supplied by the pool system. Therefore. (1.911 x 6010) I heart transfered by the pool to melting the 1.911 mol of ice. T = Temp. of the pool = /27.2th (-1.911 x 6010) J (27.2 + 273.15) 273.15 €

- As the surrounding (pool) is very large in comparisn to the system : the heat gain of lost by the system is very low. 1. Therefore Temperature of the pool assumed as const. . . © Therefore, Overall as = ( 48jce + 4Spool) = [42.049 +(~38.241)) 3/ = 3.808 JIK.

- As the surrounding (pool) is very large in comparisn to the system : the heat gain of lost by the system is very low. 1. Therefore Temperature of the pool assumed as const. . . © Therefore, Overall as = ( 48jce + 4Spool) = [42.049 +(~38.241)) 3/ = 3.808 JIK.