Question

Some water is placed in a coffee cup calorimeter. When1.0 g of an ionic solid is added, the temperature of the solution increases from 21.5°C to 24.2°C as the solid dissolves. For the dissolving process, what are the signs for${\text{ΔS}}_{sys},\Delta S_{surr},and\Delta S_{univ}$ ?

Short answer

The signsfor$\Delta S_{sys}$ is positive.

$\Delta S_{surr}=0$

$\Delta S_{univ}=0$

Step-by-step solution

Introduction to the Concept

According to the second rule of thermodynamics, the entropy of the universe increases in a spontaneous process.

Determination of the signs for ΔSsys,ΔSsurr,andΔSuniv

When 1.0 g of an ionic solid is added to the water in a coffee cup calorimeter, the temperature rises from 21.5°C to 24.2°C.

The formula $ds=\frac{dq}{T}$ could be used toexpress the change in entropy.

However, $dq=ms\Delta T$

S is a specific heat, and it has always been positive and $dq$is always dependent on$\Delta T$.

Since the system temperature rises from 21.5°C to 24.2°C,

$\Delta T$is a positive number.

Then$dq$is a positive number.

Therefore,$\Delta S_{sys}$is positive, and$\Delta S_{surr}$has no effect; thus, it is0, and$\Delta S_{univ}=0$ .