Assume that the enthalpy of fusion of ice is 6020 J/mol and does not vary apprec

Question

Assume that the enthalpy of fusion of ice is 6020 J/mol and does not vary appreciably over the temperature range 270-290 K. If one mole of ice at 0 degree C is melted by heat supplied from surroundings at 276 K, what is the entropy change in the surroundings, in J/K? A 22.1 B 21.8 C 0.0 D -21.8 E -22.1 If the change in entropy of the surroundings for a process at 431 K and constant pressure is -326 J/K, what the heat flow absorbed by for the system? A 326 kJ B 1.32 kJ C 141 kJ D 105 kJ E 141 kJ The heat of vaporization for 1.0 mole of water at 100 degree C and 1.0 atm is 40.56 kJ/mol.

Explanation / Answer

17]

S = q/T
System takes (6020J/mol)*(1 mol) = 6020 J of energy to melt 1 mole of ice at 0C. If this energy is supplied by the surroundings then the surroundings had a q of -6020 J
The entropy change of the surroundings, we we assume are at a constant temperature of 276 K is then:

S = (6020 J)/( 276 K K) = 21.8 J/K

18]

S of the surroundings of a system is -326 J/K at a constant pressure and temperature T = 451K, so q for the surroundings is:

T*S = q = (431 K)*(-326 J/K) = -141J

Negative, quantity indicating that the surroundings lost thermal energy, therefore the system in question gained 141 J of thermal energy.

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