An ice cube of volume 13.7 cm3 is initially at a temperature of -15.3 Solution D

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Explanation / Answer

Density of ice = 0.92 g/cm^3 => mass of ice = 11.8 * 0.92 g = 10.856 g Heat needed to raise temperature of 1 g of ice from - 20.4° C to 0° C = (0.5) * (20.4) cal = 10.2 cal. Heat needed to melt 1 g of ice to water at 0° C = 80 cal. Heat needed to raise the temperature of water from 0° C to 100° C = 100 cal Heat needed to convert 1 g of water at 100° C to steam = 540 cal Total heat needed to convert 1 g of ice at - 20.4° C to steam = 10.2 + 80 + 100 + 540 = 730.2 cal Heat needed to convert 10.856 g (11.8 cm^3) of ice at - 20.4° C to steam = 10.856 * 730.2 cal = 7927 cal.

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