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# Heat required to convert one gram of ice at 0 Celsius into steam at 100 Celsius

Asked by takshitashu46 9th February 2022, 6:01 PM
Heat Q1 required to convert 1 g ice at 0o C to water at 0o C  is given as

Q1 = m × Lf = 10-3 × 336 × 103 = 336 J  ........................... (1)

where m is mass of ice , Lf is latent heat of fusion of ice .

Heat Q2 required to convert 1 g water at 0o C to water at 100o C  is given as

Q2 = m × Cp × 100  = 10-3 × 4186 × 100  ≈ 419 J  ...................... (2)

Heat Q3 required to convert 1 g water at 100o C to steam at 100o C  is given as

Q3 = m × Lv = 10-3 × 2258 × 103  = 2258 J

Hence, total heat = Q1 + Q2 + Q3 = ( 336 + 419 + 2258 ) J = 3013 J
Answered by Expert 10th February 2022, 12:12 AM
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Tags: heat-energy