Thermodynamics Problem: Honey and Hot Water

Problem:

You have 100 mL of boiling water at 100°C (373.15 K). When one tablespoon of honey (approximately 21 grams) is added to this water, the system reaches thermal equilibrium, and the temperature of the water decreases slightly. Calculate the change in entropy of the water as the temperature drops from 100°C to 99°C (372.15 K). Assume that the specific heat capacity of water is 4.18 J/g°C and that the mass of the water is 100 grams (density of water is approximately 1 g/mL).

Solution:

To calculate the change in entropy (ΔS), we can use the formula:

ΔS = m \times c \times \ln (\frac{Tf}{Ti})

Where:

ΔS is the change in entropy (in J/K)
m is the mass of the water (100 grams)
c is the specific heat capacity of water (4.18 J/g°C)
T_i is the initial temperature (373.15 K or 100°C)
T_f is the final temperature (372.15 K or 99°C)

Plugging in the values:

ΔS = 100 \times 4.18 \times \ln (\frac{372.15}{373.15})

ΔS \approx 100 \times 4.18 \times -0.00268

ΔS \approx -1.12 J/K

The negative sign indicates that the entropy of the water decreases as it cools down slightly from 100°C to 99°C. This makes sense because the system (water) loses energy to the surroundings as heat, and the temperature drop corresponds to a decrease in the randomness (or entropy) of the water molecules.