Final answer:
The change in entropy when 0.350 kg of ice is melted at 0°C is calculated using the formula ΔS = Q/T, resulting in a change in entropy of approximately 428.2 J/K.
Step-by-step explanation:
The student is asking about the change in entropy when melting ice into water at the same temperature. To find the change in entropy (ΔS) of the water when melting ice, we use the formula ΔS = Q/T, where Q is the heat added to the ice, and T is the absolute temperature at which the phase change occurs.
Since the latent heat of fusion L_f for water is 334 kJ/kg, the amount of heat Q required to melt 0.350 kg of ice is calculated as Q = m⋅L_f = (0.350 kg)⋅(334 kJ/kg) = 116.9 kJ. Remember to convert this to joules, which gives Q = 116,900 J because 1 kJ = 1000 J.
The melting temperature of ice is 0°C, which corresponds to T = 273 K (since we need to use the Kelvin scale for thermodynamic calculations). Thus, the change in entropy is ΔS = Q/T = 116,900 J / 273 K = 428.2 J/K.
The change in entropy of the water can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the ice, and T is the temperature in Kelvin.
In this case, we are given the heat added to the ice, which is equal to the latent heat of fusion of water multiplied by the mass of the ice. The latent heat of fusion of water is 334 kJ/kg, so the heat added is equal to 3.34 × 10^5 J.
The temperature at which the change in entropy occurs is the melting temperature of ice, which is 0°C or 273 K. Plugging these values into the equation, we get:
ΔS = 3.34 × 10^5 J / 273 K = 1.22 × 10^3 J/K