Question

Suppose you put a cube of ice of mass 100 g into a glass of water at just above 0 degrees C. When the ice melts, about 33 kJ of energy is absorbed by the sample from the surroundings as heat. What is the change in entropy of the surroundings?

Answer 1

The change in entropy of the surroundings when 100 g of ice melts at 0°C is approximately 122 J/K, calculated using the formula for entropy change and the latent heat of fusion for water.

Step-by-step explanation:

The change in entropy of the surroundings when a cube of ice melts can be calculated using the formula ΔS = q/T, where q is the heat absorbed and T is the temperature at which the melting process occurs. Given that 33 kJ of energy is absorbed for the melting of 100 g of ice, and using the melting temperature of ice which is 273 K (0°C), the change in entropy of the surroundings can be calculated as:

ΔS = q/T = (33×1000 J) / 273 K = 120,879 J/K

However, since only 100 g (0.1 kg) of ice is melting, we will use the given value of the latent heat of fusion for water (334 kJ/kg) to find the exact amount of heat absorbed (q):

q = (0.1 kg)(334 kJ/kg) = 33.4 kJ = 33,400 J

Now, we can calculate the change in entropy for the surroundings:

ΔSsurroundings = q/T = 33,400 J / 273 K ≈ 122.344 J/K

Therefore, when 100 g of ice melts at 0°C, the change in entropy of the surroundings is approximately 122 J/K.