Lesson 5: Heat

Rule 1: Latent heat of fusion and vapourization — While ice melts at 0 °C or water boils at 100 °C, heat is absorbed but the temperature stays constant, because the heat is used to break the bonds between particles. Heat absorbed during melting is the latent heat of fusion; during boiling it is the latent heat of vapourization.

In the figure : In the figure temperature stays constant while the state changes.

• Easy : A mixture of ice and water stays at 0 °C while the ice is melting, even in a warm room.
• Medium : Boiling water in an open pan stays at 100 °C; turning up the flame only makes it boil faster, not hotter.
• Exam-type : Steam at 100 °C burns more than boiling water at 100 °C because every kilogram of steam carries an extra 2256 kJ of latent heat that it gives to the skin when it condenses.

Rule 2: Specific latent heat — Specific latent heat (विशिष्ट सुप्त उष्णता) is the heat absorbed by unit mass at constant temperature during a change of state: Q = m L. For water and ice: fusion 333 kJ/kg (80 cal/g); vapourization 2256 kJ/kg (540 cal/g). The values depend on atmospheric pressure.

• Easy : Melting 2 kg of ice at 0 °C: Q = 2 × 333 = 666 kJ.
• Medium : Turning 50 g of boiling water at 100 °C into steam: Q = 50 × 540 = 27,000 cal.
• Exam-type : Heat needed to turn 1 kg of ice at 0 °C into steam at 100 °C: melt 333 kJ + warm water (1 × 4.18 × 100 = 418 kJ) + vapourize 2256 kJ = 3007 kJ.

Spot the difference

Rule 1: Regelation — Pressure lowers the melting point of ice below 0 °C, so ice melts under pressure and re-freezes when the pressure is removed. A weighted wire slowly passes through an ice slab without cutting it into two pieces. An ice-ball is made by pressing snow.

In the figure : In the figure regelation of ice.

• Easy : Squeezing a snowball melts a thin layer of snow, which freezes again and binds the snow.
• Medium : A skater glides on a thin film of water made by the pressure of the blade.
• Exam-type : In the wire experiment the wire passes through the block and the block is still one piece, because the water above the wire freezes again.

Rule 2: Anomalous behaviour of water — From 0 °C to 4 °C water contracts on heating, so its volume is minimum and its density is maximum at 4 °C. This can be studied with Hope's apparatus. Because of this, aquatic life survives under frozen lakes. But water expands on freezing, so pipes and rocks crack in cold regions.

In the figure : In the figure anomalous expansion of water.

• Easy : Cold water below 4 °C is lighter, so it rises and freezes at the top of the lake.
• Medium : The bottom of a frozen lake stays near 4 °C, where the water is densest, so fish survive.
• Exam-type : A glass bottle full of water kept in a freezer cracks because water expands by about 9% when it turns into ice.

Spot the difference

Rule 1: Saturated air and dew point — Air is saturated when it holds the maximum water vapour possible at that temperature: 49 g/kg at 40 °C and 14.7 g/kg at 20 °C. The temperature at which unsaturated air becomes saturated on cooling is the dew point (दवबिंदू).

In the figure : In the figure saturation depends on temperature.

• Easy : Warm air can hold more vapour than cold air: 49 g/kg at 40 °C against 14.7 g/kg at 20 °C.
• Medium : Air at 40 °C that holds 14.7 g/kg of vapour is cooled; at 20 °C it becomes saturated, so its dew point is 20 °C.
• Exam-type : Dew forms on a cold glass of water when the glass is below the dew point of the air around it; the vapour condenses on it.

Rule 2: Absolute and relative humidity — Absolute humidity = mass of vapour per unit volume of air (kg/m³). Relative humidity = (actual mass of vapour ÷ mass needed for saturation at that temperature) × 100. At the dew point it is 100%. Above 60% the air feels humid; below 60% it feels dry.

• Easy : Air at 20 °C holds 7.35 g/kg of vapour: relative humidity = 7.35 ÷ 14.7 × 100 = 50%.
• Medium : Air at 40 °C holding 24.5 g/kg: RH = 24.5 ÷ 49 × 100 = 50% too, though it holds more vapour; relative humidity depends on the temperature.
• Exam-type : A room at 20 °C has air with 14.7 g/kg of vapour. Its RH is 100% and it is at the dew point, so water drops can form on a cold wall.

Spot the difference

Rule 1: Units of heat — The units of heat are the joule (SI) and the calorie. 1 cal is the heat needed to warm 1 g of water by 1 °C, from 14.5 °C to 15.5 °C. 1 kcal = 1000 cal, and 1 cal = 4.18 J. The joule is named after James Prescott Joule.

• Easy : 3 cal = 3 × 4.18 = 12.54 J.
• Medium : 2 kcal = 2000 cal = 2000 × 4.18 = 8360 J.
• Exam-type : A food packet says 250 kcal. In joules: 250 × 1000 × 4.18 = 1,045,000 J = 1045 kJ.

Rule 2: Q = m c ΔT — Heat absorbed or lost Q = m c ΔT. SI unit of c is J/kg °C; CGS unit is cal/g °C. Water has c = 1.0 cal/g °C, which is high.

In the figure : In the figure heat needed to warm water.

• Easy : Heat 200 g of water from 20 °C to 70 °C: Q = 200 × 1 × 50 = 10,000 cal.
• Medium : The same in joules: 10,000 × 4.18 = 41,800 J, or 10 kcal.
• Exam-type : 500 g of copper (c = 0.09 cal/g °C) is heated by 50 °C: Q = 500 × 0.09 × 50 = 2250 cal, much less than water would need (500 × 1 × 50 = 25,000 cal).

Spot the difference

Rule 1: Principle and calorimeter — In an isolated system, heat lost by the hot object equals heat gained by the cold object. In the mixing method with a calorimeter (उष्णतामापक), heat lost by the hot solid = heat gained by the water + heat gained by the calorimeter.

In the figure : In the figure heat lost equals heat gained.

• Easy : Mix 100 g of water at 80 °C with 100 g of water at 20 °C: the final temperature is the middle value, 50 °C.
• Medium : Mix 200 g of water at 60 °C with 300 g at 20 °C: 200 × (60 − T) = 300 × (T − 20), so 12,000 − 200T = 300T − 6000; 18,000 = 500T; T = 36 °C.
• Exam-type : A 150 g metal ball (c = 0.2 cal/g °C) at 100 °C goes into 300 g of water at 20 °C: 150 × 0.2 × (100 − T) = 300 × 1 × (T − 20); 3000 − 30T = 300T − 6000; 9000 = 330T; T ≈ 27.3 °C.

Spot the difference