Solve Increase in Entropy: 2.1x107kg Ice & Snow Avalanche

Thread starter S1CkFiSh
Start date Oct 12, 2009
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Entropy increase

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To calculate the increase in entropy for a 2.1 x 10^7 kg ice and snow avalanche sliding 0.55 km down a mountain, the relevant equation is dS = dQ/T, where dQ equals the work done, dW, calculated as mgh. The gravitational potential energy (mgh) must be determined using the mass, height, and gravitational acceleration. The temperature of the avalanche is given as 265 K, which will be used in the entropy calculation. Assistance in deriving the final entropy value is requested, emphasizing the importance of the correct application of these equations.

Oct 12, 2009

S1CkFiSh

2.1 x107 kg of ice and snow at 265 K slides 0.55 km (vertical distance) down a mountain in an avalanche. What is the increase in entropy? I am having trouble finding an equation on how to solve this problem. If anyone could help me at all, thank you.

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Oct 12, 2009

rl.bhat

Homework Helper

Change in entropy dS = dQ/T.
Here dQ = dW = mgh.

Oct 12, 2009

S1CkFiSh

Thank you very much, its very appreciated.

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