Question

Calculate the change in entropy of 144 g ice at -5 oC as it is heated to 0 oC , melted, heated to 100 oC, and then vaporized at that temperature. Suppose that the changes are brought by a heater that supplies energy at constant rate, and sketch a graph showing (a) the change in temperature of the system, (b) the enthalpy of the system, and (c) the entropy of the system as a function of time.

Answer 1

heat absorbed during change of temp of ice from -5⁰ to 0⁰= mcpdt = tds = 144 * 2.03 * 5 = 1461.6 j

Δs = m*cp*ln tf/ti = 144 * 2.03 * ln 273/268 = 5.4034 j/ ⁰C

heat added to melt ice = 336*144 = 48384 j

entropy change during melting of ice to water= Δs = 336 /273 = 1.2307 j/g ⁰C = 1.2307*144 j ⁰C = 177.230 j/⁰C

heat absorbed during change of temp of water from 0⁰ to 100⁰ = mcpdt = tds = 144 * 4.186 * 100 = 60278.4 j

Δs = m*cp*ln tf/ti = 144 *4.186 * ln 373/273 = 188.1328 j/ ⁰C

heat added to vaporize water = 2257*144 = 325008 j

entropy change during vaporization of water = Δs = 2257/373 = 6.051  j /g⁰C = 6.051 *144 ⁰C = 871.344 ⁰C

total enthalpy change during the process = 1461.6 + 48384 + 60278.4 + 325008 = 435 kJ

total change in entropy = 5.4034 + 177.230 + 188.1328 + 871.344 = 1242.1102 j /⁰C

a.

b.

c.