The entropy change for the process H2O(s) ⟶ H2O(l) It is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

Solution:

We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔS_univ is positive, then the process is spontaneous.

At both temperatures, ΔS_sys = 22. 1 J/K and q_surr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

ΔS_univ = ΔS_sys + ΔS_surr

= ΔS_sys + (q_surr/T)

= 22.1J/K + [(−6.00×103J)/263.15]K

= −0.7J/K

S_univ < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

ΔS_univ = ΔS_sys + (q_surr/T)

= 22.1J/K + [(−6.00×103J)/283.15]

K= +0.9J/K

S_univ > 0, so melting is spontaneous at 10.00 °C.