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A 15.0-kg block of ice at 0.00 °c falls into lake superior, which is a freshwater lake. the lake water is at 10.0 °c, and the latent heat of fusion for ice is 3.34 × 105 j/kg. f…
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A 15.0-kg block of ice at 0.00 °c falls into lake superior, which is a freshwater lake. the lake water is at 10.0 °c, and the latent heat of fusion for ice is 3.34 × 105 j/kg. f…
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Apr 2, 2018
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A 15.0-kg block of ice at 0.00 °c falls into lake superior, which is a freshwater lake. the lake water is at 10.0 °c, and the latent heat of fusion for ice is 3.34 × 105 j/kg. find the various changes in entropy due to the melting of the ice.
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We are given:
mass of ice = 15 kg
Ti = 0 degrees Celsius
Tf = 10 degrees Celsius
Latent heat of fusion for ice = 3..34x10^5 J/kg
The formula for entropy:
S = mH /(Tf - Ti)
S = 15 kg * 3.34x10^5 J/kg / (10-0)
S = 501 kg/K
Mirian
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Apr 7, 2018
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