Final answer:
The entropy change (ΔS) for melting 69.0 kg of ice into water at 0°C is roughly 84,382 J/K for the ice. Heat from surrounding water at 15°C is used, but the entropy decrease in the surroundings is equal and opposite to the increase in the ice, making the overall entropy change of the universe zero for this process.
Step-by-step explanation:
The task involves calculating the entropy change of the universe as a result of melting ice at 0 degrees Celsius into water at the same temperature.
The mass of the ice is given as 69.0 kg, and the latent heat of fusion of water is 334,000 J/kg, indicating the amount of heat required to melt 1 kg of ice into 1 kg of water without changing the temperature.
To calculate the change in entropy (ΔS) for the ice melting, we use the formula ΔS = Q/T, where Q is the heat added and T is the melting temperature in Kelvin:
Q = m * Lf = (69.0 kg) * (334,000 J/kg) = 23,046,000 J.
Since the melting temperature of ice is 0°C, which corresponds to 273 K, the entropy change for the ice is:
ΔS(ice) = Q / T
= 23,046,000 J / 273 K ≈
84,382 J/K.
The heat that melts the ice comes from the surrounding water at 15 degrees Celsius.
However, since it is a phase change happening at a constant temperature, the entropy change for the surrounding water needs to take into account only the change in entropy due to the transfer of heat.
That would be a decrease in entropy, which is the negative of the entropy change of the melting ice, -84,382 J/K.
Therefore, the entropy change of the universe is the sum of the entropy change of the system (the melting ice) and the entropy change of the surroundings (the water).
Since the sign difference, the changes would cancel out, leading to an overall entropy change of the universe being 0, assuming we only consider the ice and the water it melts into without other external factors.