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Grade: 11

                        

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

10 months ago

Answers : (1)

Arun
24737 Points
							
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
 
 
10 months ago
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