Answer :
The lake's entropy increases by approximately 139.56 J/K due to the boulder's impact.
To calculate the entropy change of the lake water due to the falling boulder, we can use the formula:
[tex]\[\Delta S = \frac{Q}{T}\][/tex]
Where:
- [tex]\( \Delta S \)[/tex] is the change in entropy
- [tex]\( Q \)[/tex] is the heat transferred to the lake water
- [tex]\( T \)[/tex] is the temperature of the lake water
First, let's calculate the heat transferred to the lake water [tex](\( Q \))[/tex]. The potential energy of the boulder is converted into heat energy upon impact with the water. The potential energy is given by:
- [tex]\[PE = mgh\][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the boulder (218 kg)
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\(9.81 \, \text{m/s}^2\))[/tex]
- [tex]\( h \)[/tex] is the height from which the boulder falls (19 m)
So,
- [tex]\[PE = (218 \, \text{kg})(9.81 \, \text{m/s}^2)(19 \, \text{m})[/tex]
[tex]= 40402.62 \, \text{J}\][/tex]
This energy is transferred to the lake water as heat.
Now, we'll calculate the entropy change [tex](\( \Delta S \))[/tex]. We know that the temperature of the lake water [tex](\( T \))[/tex] is given as [tex]\(16.4°C\)[/tex], which we'll need to convert to Kelvin [tex](\( K \))[/tex] because temperature must be in Kelvin for entropy calculations.
- [tex]\[T = 16.4°C + 273.15 = 289.55 \, K\][/tex]
Now, we can plug the values into the formula for entropy change:
- [tex]\[\Delta S = \frac{Q}{T} \\[/tex]
[tex]= \frac{40402.62 \, \text{J}}{289.55 \, \text{K}}[/tex]
[tex]\approx 139.56 \, \text{J/K}\][/tex]
So, the entropy change of the lake water due to the falling boulder is approximately [tex]\(139.56 \, \text{J/K}\)[/tex].
