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44) What is the change in entropy of the ice when a 0.050-kg ice cube at 0°C melts? The latent heat of fusion for water is 80 cal/g.

44) What is the change in entropy of the ice when a 0.050-kg ice cube at 0°C melts? The latent heat of fusion for water is 80 cal/g.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Please show all calculations and type them out So that I will be able to read them and understand what you're saying I have bad eyesight and it is very difficult for me to read handwriting 

**Entropy Change During Melting of Ice** **Question 44:** What is the change in entropy of the ice when a 0.050-kg ice cube at 0°C melts? The latent heat of fusion for water is 80 cal/g. --- **Explanation:** To solve this problem, we need to calculate the change in entropy (ΔS) of the ice as it melts. **Given:** - Mass of ice (m) = 0.050 kg - Latent heat of fusion for water (L) = 80 cal/g First, convert the mass of ice from kilograms to grams: \[ 0.050 \text{ kg} = 50 \text{ g} \] Next, calculate the total amount of heat (Q) absorbed during melting: \[ Q = m \times L \] \[ Q = 50 \text{ g} \times 80 \text{ cal/g} \] \[ Q = 4000 \text{ cal} \] The change in entropy (ΔS) is given by the formula: \[ \Delta S = \frac{Q}{T} \] Where: - \( Q \) is the heat absorbed - \( T \) is the absolute temperature in Kelvin (K) The melting point of ice is 0°C, which is 273.15 K. Now, calculate ΔS: \[ \Delta S = \frac{4000 \text{ cal}}{273.15 \text{ K}} \] \[ \Delta S \approx 14.64 \text{ cal/K} \] Thus, the change in entropy of the ice when it melts is approximately **14.64 cal/K**.
Transcribed Image Text:**Entropy Change During Melting of Ice** **Question 44:** What is the change in entropy of the ice when a 0.050-kg ice cube at 0°C melts? The latent heat of fusion for water is 80 cal/g. --- **Explanation:** To solve this problem, we need to calculate the change in entropy (ΔS) of the ice as it melts. **Given:** - Mass of ice (m) = 0.050 kg - Latent heat of fusion for water (L) = 80 cal/g First, convert the mass of ice from kilograms to grams: \[ 0.050 \text{ kg} = 50 \text{ g} \] Next, calculate the total amount of heat (Q) absorbed during melting: \[ Q = m \times L \] \[ Q = 50 \text{ g} \times 80 \text{ cal/g} \] \[ Q = 4000 \text{ cal} \] The change in entropy (ΔS) is given by the formula: \[ \Delta S = \frac{Q}{T} \] Where: - \( Q \) is the heat absorbed - \( T \) is the absolute temperature in Kelvin (K) The melting point of ice is 0°C, which is 273.15 K. Now, calculate ΔS: \[ \Delta S = \frac{4000 \text{ cal}}{273.15 \text{ K}} \] \[ \Delta S \approx 14.64 \text{ cal/K} \] Thus, the change in entropy of the ice when it melts is approximately **14.64 cal/K**.
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