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irina1246 [14]
3 years ago
9

A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re

ached. Calculate the entropy change of the leadthat occurs during this process.
Chemistry
1 answer:
sveticcg [70]3 years ago
6 0

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula  -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass  = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature  10.0°C + 273 = 283 K

T₁ = initial temperature ,  40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ )  = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

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