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Specific Latent Heat
Heating Curve - Latent Heat
- T1 is the melting point whereas T2 is the boiling point.
- From Q to R and S to T, the temperature remains constant because the heat supplied to the object is used to overcome the forces of attraction that hold the particles together.
- Heat obsorbs during Q-R is called the latent heat of fusion.
- Heat obsorbs during S-T is called the latent heat of vaporisation.
Cooling Curve - Latent Heat
- T1 is the condensation point, T2 is the freezing point whereas T3 is room temperature.
- During Q-R and S-T, the temperature remains unchanged. This is because the energy produced during the formation of bonds is equal to the heat energy released to the surroundings during cooling.
- The heat energy released during Q-R is called the latent heat of vaporization.
- The heat energy released during S-T is called the latent heat of fusion.
Answer:
There are 2 processes involve when an ice is converted into water at 25°C.
Ice at 0°C ------> Water at 0°C -------> Water at 25°C
Energy absorbed to convert 0.5kg from Ice at 0°C to Water at 0°C
Q1 = mL = (0.5)(334000) = 167000J
Energy absorbed to convert 0.5kg from watyer at 0°C to Water at 25°C
Q2 = mcθ = (0.5)(4200)(25-0) = 52500J
Total energy required = Q1 + Q2 = 167000 + 52500 = 219500J
Answer:
The processes involved:
Ice (-20°C) -----> Ice (0°C) -----> Water (0°C) -----> Water (100°C)
-----> Steam (100°C) -----> Water (120°C)
Energy required:
Ice (-20°C) to Ice (0°C), Q1 = mcθ = (0.001)(2100)(20) = 42J
Ice (0°C) to Water (0°C), Q2 = mL = (0.001)(334000) = 334J
Water (0°C) to Water (100°C), Q3 = mcθ = (0.001)(4200)(100) = 420J
Water (100°C) to Steam (100°C), Q4 = mL = (0.001)(2260000) = 2260J
Steam (100°C) to Steam (120°C), Q5 = mcθ = (0.001)(2020)(20) = 40.4J
Total energy required
= Q1 + Q2 + Q3 + Q4 + Q5
= 42 + 334 + 420 + 2260 + 40.4 = 3096.4J
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