Question

Ice cubes at 0 c are poured into a glass that contains 150 g of liquid water at 20 c. The contents of the glass are gently stirred and after short time the ice melts and the liquid cools to 0 c. No appreciable heat exchange with the environment takes place.

a) calculate the entropy change of the universe (system plus environment) during the process. Cp=Cv = 4.18 kj/kgK
b)if the same process were performed reversibly, how much useful work would be obtained from it?

Answer 1

Total Entropy Change

A system and surrounding are both components of the universe (total), which can be seen in this equation:

?Stotal=?Suniv= ?Ssys + ?Ssurr>0

the system and surroundings collective represent the change in entropy. The entropy change of the surroundings is driven by heat flow and the heat flow determines the sign of ?Ssurr. Therefore if we had an exothermic reactions with a constant temperature the system would cause heat to flow into the surroundings, which causes ?Ssurr to become positive. The opposite can be see in an endothermic reaction.

Entropy change for the System

We begin by first calculating the difference between the standard entropy values of the products and standard entropy values of the reactants:

?Sreaction=?npSproducts - ?npSreactants

where S represents entropy, and np and nr represent moles of products and reactions. Once the you have calculated the entropy of the system, you can calculate the entropy change for the surroundings.

Entropy change for the Surroundings

And from the Second Law of Thermodynamics, we know that the sign of ?Stotal determines whether the reaction under investigation happens spontaneously. For calculating the entropy for the surrounding, we can use the general definition of S:

?Ssurroundings = ? dS surroundings = ? (?q surroundings / T surroundings)

In which ?q is the element of heat and T is temperature (in K). From the conservation of energy, q surroundings and q system are related:

q surroundings = - q system

In general, to do the integration, one needs to know T along the path. As a common specific case, if the surrounding is big enough (e.g. the lab or the universe, with respect to the beaker in which the experiment is being done is too big), one can assume that temperature (T) does not change during the course of the reaction, in which case the integral is simplified:

?Ssurroundings = q surroundings / T surroundings = - q system / T surroundings

in which the value for q system is the same as enthalpy change and internal energy (internal energy change) of the system at constant pressure and constant volume, respectively.

Example

For example, if one has a generic reaction

Reactants --> Products qrxn = 75 kJ at 300K

Then, for the surrounding, one has,

?Ssurroundings = - (75