The amount of heat required to convert 5 gm of ice at −200C to water at 1000C is
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The amount of heat required to convert 1 gm of Ice at 0∘ to 1 gm of steam at 100∘C.
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The amount of heat required to convert 1 gm of ice at −100C to steam at 1000C is
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Answer these: [4 Mark]
a) How much energy is transferred when 1 gm of boiling water at 100 °C condenses to water at 100 °C? Ans: When 1 g of boiling water at 100 °C condenses to water at 100 °C there is no change in temperature and the energy transferred is given by the latent heat of vaporisation of water which is equal to 2260 Joule.
b) How much energy is transferred when 1 gm of boiling water at 100 °C cools to water at 0 °C? Ans:Stage 1 → Heat releases from water vapour at 100°C
Stage 2 → Heat energy releases from water at 100°C
When converts into water at 0°C
Q2 = mSΔT,
Here ΔT = 100°C – 0°C = 100°C
(∵ m = 1 gm; S = 1 Cal/gm°C.)
= 1 ×1 ×100 = 100 Cal.
Total energy releases
⇒ Q1 + Q2 = 540 + 100 = 640 Cal.
c) How much energy is released or absorbed when 1 gm of water at 0 °C freezes to ice at 0 °C? Ans:The amount of energy released when 1 gm of water at 0°C freezes to ice is same as when the same amount of ice is melted into water i.e, each 1g will release heat energy equal to 334 J on freezing as latent heat of Fusion of ice is L = 334 J/g.
d) How much energy is released or absorbed when 1 gm of steam at 100 °C turns to ice at 0 °C? Ans:
Latent heat of vaporization = 540cal/g
Energy transferred when 1g of boiling water at 100°C cools to water at 0°C is - 100cal
Latent heat change on changing the state from water to ice = 80cal/g
Hence, energy released when 1gm of steam at 100°C turns to ice at 0°C = 540 + 100 + 80 = 720cal.
A total of 334 J of energy are required to melt 1 g of ice at 0°C, which is called the latent heat of melting. At 0°C, liquid water has 334 J g−1 more energy than ice at the same temperature.
CHANGES OF STATE OF WATER. - The change from solid to liquid is called fusion, or melting. - To melt 1 gram of ice requires 80 calories. (A calorie is defined as the amount of energy needed to raise one gram of water 1°C.)
It takes 100 calories to heat 1 g. water from 0˚, the freezing point of water, to 100˚ C, the boiling point. However, 540 calories of energy are required to convert that 1 g of water at 100˚ C to 1 g of water vapor at 100˚ C. This is called the latent heat of vaporization.
To change one gram of 0°C ice to 100°C steam, we need to consider the different phase changes and the associated heat transfer. The total amount of heat required is approximately 7218.6 calories. The correct answer choice is option d: 720.
The enthalpy of vaporization of water is 2256.4J/g, meaning that 2256.4J of heat are required to convert 1g of water at 100C to steam at 100C. In this case, we have 100g of water at 100C and we want to vaporize it to steam (also at 100C). This requires ΔH = 2256.4(100) = 225,640J or 225.64kJ of heat energy.
Introduction: My name is Nicola Considine CPA, I am a determined, witty, powerful, brainy, open, smiling, proud person who loves writing and wants to share my knowledge and understanding with you.
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