Since L_ice = 333 J/g, the change in entropy (in cal/K) when 1.00 kg of ice melts is A) 144 B) 291 C) 312 D) 516 E) 80

Question

Quizplus · Accepted Answer

The change in entropy ($\Delta S$) can be calculated using the formula $\Delta S = \frac{q}{T}$, where $q$ is the heat absorbed or released (in joules) and $T$ is the temperature in Kelvin. Given that the latent heat of fusion for ice is 333 J/g, and knowing that 1.00 kg of ice is equivalent to 1000 g, the total heat ($q$) involved in melting 1.00 kg of ice is $333 \times 1000 = 333,000$ J. To convert joules to calories, we use the conversion factor $1 \text{ cal} = 4.184 \text{ J}$, so $333,000 \text{ J} = \frac{333,000}{4.184} \approx 79,600 \text{ cal}$. The melting point of ice is 0°C, which is 273.15 K. Therefore, $\Delta S = \frac{79,600}{273.15} \approx 291 \text{ cal/K}$.

Since Lice = 333 J/g, the Change in Entropy (In

Since L_ice = 333 J/g, the Change in Entropy (In