An avalanche of ice and snow of mass 2,000 kg slides a vertical distance of 180 m down a mountainside.If the temperature of the ice,snow,mountain and surrounding air are all at 0°C,what is the change in entropy of the universe?

A) 25,846 J/K
B) 12,923 J/K
C) 6,462 J/K
D) 4308 J/K
E) 1,319 J/K

Question

An avalanche of ice and snow of mass 2,000 kg slides a vertical distance of 180 m down a mountainside.If the temperature of the ice,snow,mountain and surrounding air are all at 0°C,what is the change in entropy of the universe?
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A) 25,846 J/K
B) 12,923 J/K
C) 6,462 J/K
D) 4308 J/K
E) 1,319 J/K

Quizplus · Accepted Answer

To calculate the change in entropy of the universe, we need to use the formula: ΔS_univ = ΔS_system + ΔS_surroundings. Here, the system is the avalanche of ice and snow, and the surroundings are the mountain and the air around it.

We know that the temperature of the system and the surroundings are the same, so ΔS_system = Q/T, where Q is the heat transferred and T is the temperature. Since there is no change in temperature, ΔS_system = 0.

The only change in entropy comes from the surroundings, which experience an increase in entropy as the avalanche moves down the mountain. We can calculate this using the formula: ΔS_surroundings = Q/T. Here, Q is the heat transferred from the avalanche to the surroundings, and we can calculate it using the formula for potential energy: Q = mgh, where m is the mass of the avalanche, g is the acceleration due to gravity, and h is the height through which the avalanche falls.

Therefore, Q = (2000 kg)(9.8 m/s^2)(180 m) = 3,528,000 J.

Since the temperature is 0°C = 273 K, ΔS_surroundings = Q/T = 3,528,000 J / 273 K = 12,923 J/K.

Therefore, the answer is B.

An Avalanche of Ice and Snow of Mass 2,000 Kg