Final answer:
In an isobaric and isothermal process, q-surroundings and ∆H-surroundings are equal in magnitude but opposite in sign. The heat flow from the surroundings to the system is enthalpy change of the system at constant pressure, and the entropy change of the surroundings is directly related to that enthalpy change.
Step-by-step explanation:
The relationship between q-surroundings and ∆H-surroundings for an isobaric (constant pressure) and isothermal (constant temperature) process is that they are equal in magnitude but opposite in sign. According to thermodynamics, the heat flow qp from the surroundings to the system is equal to the enthalpy change ∆H sys of the system at constant pressure.
Moreover, the entropy change of the surroundings, ∆S surr, is related to the enthalpy change of the system; mathematically, it is given as ∆S surr = -∆H sys / T, where T is the temperature of the surroundings. This implies that when the system absorbs heat (∆H sys > 0), the surroundings get cooler due to heat loss, while if the system releases heat (∆H sys < 0), the surroundings get warmer.
When considering the sign convention, a positive ∆H indicates that the surroundings lose heat (endothermic process), making qp negative for the surroundings, while a negative ∆H indicates that the surroundings gain heat (exothermic process), making qp positive for the surroundings.