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# Someone pls do this

Asked by pukai7229 2nd November 2018, 11:00 PM
Required changes for the ice to convert from -12°C ice to 100°C steam
(1) Conversion of ice from -12°C to ice at 0 °C
(2) conversion of ice at 0 °C to water at 0 °C (latent heat of fusion)
(3) Change of water temperature from 0 °C to 100 °C
(4) Change of water from 100 °C to steam at 100 °C (latent heat of vaporaisation)

Heat energy required for Conversion of ice from -12°C to ice at 0 °C = mass×specific heat×Temperature increase
= 3 × 2.1×105 × 12 = 75.6×105 J ................................................(1)
Heat energy required for conversion of ice at 0 °C to water at 0 °C = 3×3.35×105 = 10.05×105 J ...........................................(2)

Heat energy required for Change of water temperature from 0 °C to 100 °C = 3×4.186×103 × 100  = 12.558×105 ..................(3)
Heat energy required for Change of water from 100 °C to steam at 100 °C = 3×2.256×106  = 6.768×106 J ......................(4)

By adding all heat energy from eqn.(1) to eqn.(4), we get total heat energy = 16.59 ×106 J
Answered by Expert 3rd November 2018, 3:48 AM
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Tags: heat-energy