Assignment 10.3: Find the change in the entropy of 3.00 × 102 g of lead when it melts at 327° C. Lead has a latent heat of fusion of 2.45 × 104 J/kg. (b) Suppose the same amount of energy is used to melt part of a piece of silver which is already at its melting point of 961° C. Find change in the entropy of the silver?
Solution
When 1 kg of lead is melted, 2.45 × 104 J of heat is added to it. Therefore, when 3 × 102 g = 3 × 10-1 kg of lead is melted 2.45 × 104 × 3 × 10-1 = 7.35 × 103 J of heat ΔQ is added to it. Similarly, the temperature is 327° C = 327 + 273 = 600 K. So, apply formula ΔS = ΔQ/T,
(b) Temperature of silver, T = 961° C = 961 + 273 = 1234 K
Amount of heat ΔQ = 7.35 × 103 J
Change in entropy, ΔS =?
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