It can be shown that the change in entropy in heating or cooling a sample is given by the relation ΔS = mc ln (TF/Ti) , where m is the mass of the sample, c is the specific heat of the sample, and TF and Ti are the final and initial temperatures, respectively. What is the change in entropy when 10.0 grams of lead with a specific heat of 0.452 J/g K is heated from 10.0°C to 50.0°C?
A) 0.45 J/K
B) 0.60 J/K
C) 0.90 J/K
D) 0.30 J/K
Correct Answer:
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