MHTCET 14 May 2023 Shift 1 Solved Paper

To find the entropy change of the surroundings for the given process, we can use the following thermodynamic relation which connects the enthalpy change (∆H) of a system to the entropy change of the surroundings ( ∆S‌surroundings ‌ ) at constant temperature:
∆S‌surroundings ‌‌‌=−‌

∆H

T

In this formula, ∆H is the enthalpy change for the system, and T is the temperature at which the process takes place. The negative sign indicates that if the process is endothermic (∆H>0) for the system, the entropy of the surroundings decreases ( ∆S‌surroundings ‌<0 ), and vice versa for an exothermic process.
For the reaction given:
H2O(s)→H2O(ℓ)
the enthalpy change is given as +7‌kJ‌mol−1 or +7000Jmol−1 when converted to joules, since 1‌kJ=1000J, and the temperature is 300K. Plugging these values into our equation gives:
∆S‌surroundings ‌=−‌

+7000Jmol−1

300K

To simplify this:
∆S‌surroundings ‌‌=−‌

7000

300

JK−1mol−1
∆S‌surroundings ‌‌=−23.333...JK−1mol−1
Therefore, the entropy change of the surroundings for the melting of ice at 300K with an enthalpy change of +7‌kJ‌mol−1 is approximately:
∆S‌surroundings ‌≈≈−23.3JK−1mol−1
This corresponds to Option B:−23.3JK−1.