Question related to latent heat of fusion of ice

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SUMMARY

The discussion centers on calculating the specific heat of a metal using the latent heat of fusion of ice. When 1 gram of ice at 0°C melts, it contracts by 0.091 cc, indicating a heat absorption of 80 cal/gm. In the experiment, 9 grams of metal heated to 200°C was introduced into an ice calorimeter, resulting in a volume decrease of 0.0182 cc. The calculated specific heat of the metal is determined to be 0.09 cal/gm°C, confirming the relationship between heat transfer and volume change in the calorimeter setup.

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Homework Statement



one gram of ice at 0 degree C contracts in volume by 0.091cc on melting. 9gm of a metalis heated to 200 degree C and dropped into and ice calorimeter. The decrease in volume was found to be 0.0182 cc. Calculate the Sp. heat of the metal (ANS:0.09cal/gmC)

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The Attempt at a Solution


In my opinion when 80cal/gm is supplied to ice it's volume will decrease by 0.091cc as it will turn into water. So when the Volume decreases by 0.0182 it means that heat supplied will be doubled. so 160 cal/gm = 9gm * S * 200 which which will give s = 0.08cal/gmC.
 
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