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 reached.Calculate the entropy change of the lead that occurs during this process.The specific heat of lead is 130 J/(kg • K).

Question

Quizplus · Accepted Answer

We can use the formula for entropy change:

ΔS = Q/T

where Q is the heat transferred and T is the temperature at which the transfer occurs.

The heat transferred from the lead to the water is:

Q = mcΔT

where m is the mass of the lead, c is the specific heat of lead, and ΔT is the temperature change.

Q = (2.00 kg)(130 J/(kg•K))(40.0°C - 10.0°C)
Q = 7800 J

The temperature at which the heat transfer occurs is the average of the initial and final temperatures:

T = (40.0°C + 10.0°C)/2
T = 25.0°C

Therefore, the entropy change is:

ΔS = Q/T
ΔS = 7800 J/298 K
ΔS = -26.2 J/K

The negative sign indicates that the entropy of the lead decreases during this process. Therefore, the correct choice is (C) -26.2 J/K.

A 200 Kg Piece of Lead at 40