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# The amount of heat required to convert 1 gm of ice at 0℃ into steam at 100℃ is

Arun
25763 Points
one year ago
m = 1 g = 0.001 kg

T1 = 0℃ and T2 = 100℃

T2 - T1 = 100 ℃

As The heat required to melt 1g of ice to water can be calculated as

q = Lm m

= 334000 (J/kg) 0.001 (kg)

= 334 (J)

= 334 (J)

The heat required to convert 1g of water at 0℃ to 1g of water at 100 ℃ can be calculated as

q = m c (T2 - T1)

= 0.001 kg * 4185.5 J/(kg⋅K) * 100 K

= 418.55 J

The heat required to convert 1g of water to steam can be calculated as

q = Lf m

= 2258000 (J/kg) 0.001 (kg)

=2258 (J)

= 2258 (J)

So, total amount of heat required

= 334 + 418.55 + 2258 J

= 3010.55 J

= 3.01055 kJ of heat