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Answer :
The entropy change of (a) the lead piece is -1957 J/°C, (b) the lake is +1957 J/°C, and (c) the universe is 0.
To determine the entropy change of the lead piece, the lake, and the universe, we can use the equation:
ΔS = m * c * ln(T2/T1)
where ΔS is the entropy change, m is the mass, c is the specific heat, T1 is the initial temperature, and T2 is the final temperature.
(a) To calculate the entropy change of the lead piece:
- Mass (m) of the lead piece is given as 5.2 kg.
- Specific heat (c) of lead is given as 128 J/(kg · °C).
- Initial temperature (T1) of the lead piece is 240°C.
- Final temperature (T2) of the lead piece is 12°C (the temperature of the lake).
Using the formula, we can calculate the entropy change of the lead piece:
ΔS = 5.2 kg * 128 J/(kg · °C) * ln(12°C/240°C)
Simplifying the calculation, we have:
ΔS = 5.2 kg * 128 J/(kg · °C) * ln(0.05)
Calculating further:
ΔS ≈ 5.2 kg * 128 J/(kg · °C) * (-2.995)
ΔS ≈ -1957 J/°C
Therefore, the entropy change of the lead piece is approximately -1957 J/°C.
(b) To calculate the entropy change of the lake:
The entropy change of the lake is equal in magnitude but opposite in sign to that of the lead piece. Therefore, the entropy change of the lake is approximately +1957 J/°C.
(c) To calculate the entropy change of the universe:
The entropy change of the universe is the sum of the entropy changes of the lead piece and the lake. Since they have equal magnitudes but opposite signs, the entropy change of the universe is zero.
Learn more about entropy change here: https://brainly.com/question/27549115
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