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$\degree$C, what is the change in entropy of the universe?

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:$$\Delta S_{avalanche} = m c_p \ln \left(\frac{T_f}{T_i}\right)$$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 -1°C. The specific heat of ice is 2.09 J/g°C. The mass of the avalanche is 2500 kg. The heat transferred from the avalanche to the surroundings is:$$Q = m c_p (T_f - T_i) = (2500 kg)(2.09 J/g°C)(0°C - (-1°C)) = 5225 J$$The temperature of the surroundings is 0°C. Therefore, the change in entropy of the surroundings is:$$\Delta S_{surroundings} = -\frac{Q}{T} = -\frac{5225 J}{273 K} = -19.1 J/K$$The change in entropy of the universe is:$$\Delta S_{universe} = \Delta S_{avalanche} + \Delta S_{surroundings} = (14000 J/K) + (-19.1 J/K) = 13980.9 J/K$$Therefore, the correct answer is A.

An Avalanche of Ice and Snow of Mass 2500 Kg °\degree°

An Avalanche of Ice and Snow of Mass 2500 Kg $\degree$