A 16-gram lead bullet slams into a concrete wall and melts on impact.If the temperature of the speeding bullet was 25 $\circ$ C and if the specific heat of lead is 0.13 kJ/(kgK)and its latent heat of fusion is 23 kJ/kg,what is the change in entropy of the bullet during this process? The melting point of lead is 601K.

Question

Quizplus · Accepted Answer

The change in entropy can be calculated using the formula ΔS = Q/T, where Q is the heat absorbed or released during the process and T is the temperature. In this case, the bullet absorbs heat equal to the product of its mass, the latent heat of fusion, and the fraction of the bullet that melts, which is (16 g) × (23 kJ/kg) × (1/16) = 23 kJ. The temperature of the bullet is given in Fahrenheit, so it needs to be converted to Kelvin: (25 − 32) × 5/9 + 273 = 268 K. Therefore, ΔS = 23 kJ / 268 K = 2.08 J/K, which is closest to choice A.

A 16-Gram Lead Bullet Slams into a Concrete Wall and Melts