22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and
latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its
temperature respectively are
A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C

Question

Vikas TU · Answer

Answer would be option C The ice would take 8000cal to turn into water by formula mass×80 Then the water would take 10000 cal by formula mass× c × change in temp(which is 100) So remaining 4320cal would turn 8g of water into vapour

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22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its temperature respectively are A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C

22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and
latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its
temperature respectively are
A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C

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22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its temperature respectively are A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C

22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its temperature respectively are A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C

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22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal g–1 and
latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its
temperature respectively are
A. 8g, 100°C B. 100g, 90°C C. 92g, 100°C D. 82g, 100°C