An avalanche of ice and snow of mass 2500 kg slides a vertical distance of 160 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) 14,000 J/K
B) 10,000 J/K
C) 1400 J/K
D) 1100 J/K

Question

Quizplus · Accepted Answer

The change in entropy of the universe is equal to the change in entropy of the avalanche plus the change in entropy of the surroundings. The change in entropy of the avalanche is given by:$$\Delta S_{avalanche} = m c_p \ln \left(\frac{T_f}{T_i}ight)$$where m is the mass of the avalanche, $c_p$ is the specific heat of ice, $T_f$ is the final temperature of the avalanche, and $T_i$ is the initial temperature of the avalanche.The change in entropy of the surroundings is given by:$$\Delta S_{surroundings} = -\frac{Q}{T}$$where Q is the heat transferred from the avalanche to the surroundings and T is the temperature of the surroundings.In this case, the final temperature of the avalanche is 0°C and the initial temperature of the avalanche is also 0°C. Therefore, the change in entropy of the avalanche is zero. The heat transferred from the avalanche to the surroundings is equal to the change in potential energy of the avalanche, which is given by:$$Q = mgh$$where g is the acceleration due to gravity and h is the vertical distance that the avalanche slides.Therefore, the change in entropy of the surroundings is:$$\Delta S_{surroundings} = -\frac{mgh}{T}$$The change in entropy of the universe is equal to the sum of the change in entropy of the avalanche and the change in entropy of the surroundings, which is:$$\Delta S_{universe} = \Delta S_{avalanche} + \Delta S_{surroundings} = -\frac{mgh}{T}$$Substituting the given values into this equation, we get:$$\Delta S_{universe} = -\frac{(2500 	ext{ kg})(9.8 	ext{ m/s}^2)(160 	ext{ m})}{273 	ext{ K}} = -14,000 	ext{ J/K}$$Therefore, the correct answer is A.

An Avalanche of Ice and Snow of Mass 2500 Kg