Chapter in a nutshell: Heat raises temperature via specific heat capacity ($Q=mc\Delta T$); when bodies mix, heat lost = heat gained (calorimetry). During a change of state, heat is absorbed/released at constant temperature — the latent heat ($Q=mL$). Water's unusually high specific heat explains many everyday effects.
1. Heat vs Temperature
- Heat is energy that flows from a hotter to a colder body; unit joule (J) (also calorie).
- Temperature is the degree of hotness; unit °C or K ($T(\text{K}) = t(°\text{C}) + 273$). Heat flow depends on temperature difference.
2. Heat Capacity & Specific Heat Capacity
- Heat capacity (C): heat required to raise a body's temperature by 1 °C (or 1 K); unit J °C⁻¹ or J K⁻¹. $C = mc$.
- Specific heat capacity (c): heat required to raise the temperature of 1 kg of a substance by 1 °C; unit J kg⁻¹ °C⁻¹.
| Substance | c (J kg⁻¹ °C⁻¹) |
|---|---|
| Water | 4200 (very high) |
| Ice | 2100 |
| Aluminium | 900 |
| Iron | 460 |
| Copper | 390 |
| Mercury | 140 |
3. Principle of Calorimetry (Method of Mixtures)
When a hot body is mixed with a cold one and no heat is lost to the surroundings: $$\textbf{Heat lost by hot body} = \textbf{Heat gained by cold body}$$ $$m_1 c_1 (T_1 - T) = m_2 c_2 (T - T_2)$$ A calorimeter (copper vessel) is used; its own heat capacity must be included in accurate work.4. Change of State & Latent Heat
During melting/boiling the temperature stays constant even as heat is supplied — the energy goes into breaking inter-molecular bonds (potential energy), not raising kinetic energy. $$\boxed{Q = m\,L}$$| Quantity | Definition | Value (water) |
|---|---|---|
| Specific latent heat of fusion | heat to change 1 kg solid → liquid at melting point | ice: $3.36\times10^5$ J kg⁻¹ (336 kJ/kg) |
| Specific latent heat of vaporisation | heat to change 1 kg liquid → gas at boiling point | water/steam: $2.26\times10^6$ J kg⁻¹ (2260 kJ/kg) |
- Evaporation (surface, any temperature) causes cooling (it takes latent heat from the surroundings) — why sweating cools us and water stays cool in an earthen pot.
5. Why Steam Burns Are Worse Than Boiling-Water Burns
Steam at 100 °C carries the extra latent heat of vaporisation (2260 kJ/kg). When it condenses on skin it releases this large latent heat plus the heat of cooling, so it scalds far more than the same mass of water at 100 °C.6. High Specific Heat Capacity of Water — consequences
- Excellent coolant (radiators of cars/engines, industrial cooling).
- Moderates climate: water bodies heat and cool slowly, so coastal places have milder weather; land/sea breezes arise because land heats/cools faster than the sea.
- Hot-water bottles stay warm for a long time; water is used in hot-water heating systems.
7. Worked Numerical Examples (ICSE pattern)
Q1. How much heat raises 2 kg of water from 20 °C to 70 °C? ($c = 4200$) Solution: $Q = mc\Delta T = 2\times4200\times50 = \mathbf{420000\ J}$.Q2. 0.5 kg of water at 80 °C is mixed with 0.5 kg at 20 °C. Find the final temperature (no losses). Solution: equal masses & c → $T=\dfrac{80+20}{2}=\mathbf{50\ °C}$.
Q3. Heat needed to melt 2 kg of ice at 0 °C. ($L_f = 3.36\times10^5$) Solution: $Q = mL = 2\times3.36\times10^5 = \mathbf{6.72\times10^5\ J}$.
Q4. Heat to convert 1 kg of water at 100 °C into steam at 100 °C. ($L_v = 2.26\times10^6$) Solution: $Q = mL = 1\times2.26\times10^6 = \mathbf{2.26\times10^6\ J}$.
Q5. Find the total heat to convert 1 kg of ice at 0 °C to water at 50 °C. ($L_f=3.36\times10^5$, $c=4200$) Solution: melt: $1\times3.36\times10^5 = 3.36\times10^5$ J; heat: $1\times4200\times50 = 2.1\times10^5$ J; total $=\mathbf{5.46\times10^5\ J}$.
Q6. 100 g of a metal at 100 °C is dropped into 200 g water at 20 °C; final temp 25 °C. Find the metal's specific heat. ($c_w=4.2$ J g⁻¹ °C⁻¹) Solution: heat gained by water $=200\times4.2\times5=4200$ J; heat lost by metal $=100\times c\times75$. Equate: $7500c=4200\Rightarrow c=\mathbf{0.56\ J\,g^{-1}\,°C^{-1}}$.
8. Key Terms — Quick Glossary
| Term | One-line definition |
|---|---|
| Heat | energy flowing due to a temperature difference; unit J. |
| Temperature | degree of hotness; unit °C / K. |
| Heat capacity | heat to raise a body's temp by 1 °C ($C=mc$). |
| Specific heat capacity | heat to raise 1 kg by 1 °C (J kg⁻¹ °C⁻¹). |
| Water equivalent | mass of water needing the same heat as the body. |
| Principle of calorimetry | heat lost = heat gained. |
| Latent heat of fusion | heat to melt 1 kg of solid at its m.p. |
| Latent heat of vaporisation | heat to vaporise 1 kg of liquid at its b.p. |
| Evaporation | surface vaporisation at any temp; causes cooling. |
9. Common Mistakes to Avoid
- Using $Q=mc\Delta T$ during melting/boiling — there ΔT = 0; use $Q=mL$ instead.
- Forgetting to add the melting + heating stages in multi-step problems.
- Mixing up heat capacity (per body) and specific heat capacity (per kg).
- Unit slips: keep mass in kg with c in J kg⁻¹ °C⁻¹ (or g with J g⁻¹ °C⁻¹) consistently.
- Forgetting the calorimeter's own heat capacity in precise calorimetry.
10. Likely Exam Questions (with crisp answers)
- Define specific heat capacity; state its SI unit. → Heat to raise 1 kg by 1 °C; J kg⁻¹ °C⁻¹.
- State the principle of calorimetry. → Heat lost by the hot body = heat gained by the cold body.
- Why does the temperature stay constant during melting/boiling? → The heat supplied is used to break inter-molecular bonds (latent heat), not to raise temperature.
- Define specific latent heat of fusion. → Heat to change 1 kg of solid to liquid at its melting point.
- Why is a steam burn more severe than a boiling-water burn? → Steam gives out extra latent heat of vaporisation on condensing.
- Why does water stay cool in an earthen pot? → Evaporation through the pores takes latent heat from the water, cooling it.
- Give two consequences of water's high specific heat capacity. → Good coolant; moderates coastal climate.
- Define water equivalent. → Mass of water requiring the same heat as the body for the same temperature rise.
- Why is water used as a coolant in radiators? → Its high specific heat lets it absorb a lot of heat for a small temperature rise.
- State the relation $Q = mL$ and name the quantities. → Heat for change of state = mass × specific latent heat.
- Why does evaporation cause cooling? → Fast molecules escape, taking latent heat from the liquid, lowering its temperature.
- Convert 27 °C to kelvin. → $27+273 = 300$ K.
11. Evaporation vs Boiling
| Evaporation | Boiling |
|---|---|
| only at the surface | throughout the bulk |
| at any temperature | at a fixed boiling point |
| slow, causes cooling | rapid, bubbles form |
12. Heating Curve of Ice → Water → Steam
As heat is supplied steadily to ice below 0 °C:- Ice warms to 0 °C ($Q=mc_{ice}\Delta T$) — temperature rises.
- Ice melts at 0 °C ($Q=mL_f$) — temperature constant (a flat step).
- Water warms 0 → 100 °C ($Q=mc_{water}\Delta T$) — rises.
- Water boils at 100 °C ($Q=mL_v$) — temperature constant (a longer flat step, since $L_v>L_f$).
- Steam superheats — rises again.
13. More Worked Numericals
Q7. Find the heat capacity of 5 kg of copper. ($c=390$ J kg⁻¹ °C⁻¹) Solution: $C=mc=5\times390=\mathbf{1950\ J\,°C^{-1}}$.Q8. A 2 kW immersion heater is switched on for 5 min in 4 kg of water at 30 °C. Find the final temperature (no loss). ($c=4200$) Solution: heat $=2000\times300=6\times10^5$ J; $\Delta T=\dfrac{Q}{mc}=\dfrac{6\times10^5}{4\times4200}=35.7$ °C → final $\approx\mathbf{65.7\ °C}$.
Q9. How much ice at 0 °C can be melted by 1.68×10⁶ J of heat? ($L_f=3.36\times10^5$) Solution: $m=Q/L=\dfrac{1.68\times10^6}{3.36\times10^5}=\mathbf{5\ kg}$.
Q10. 50 g steam at 100 °C is passed into water — how much heat does it release on just condensing? ($L_v=2.26\times10^6$ J kg⁻¹) Solution: $Q=mL=0.05\times2.26\times10^6=\mathbf{1.13\times10^5\ J}$.
14. More Exam Questions (with crisp answers)
- Distinguish evaporation and boiling. → Evaporation: surface, any temperature, cooling; boiling: bulk, fixed temperature.
- Name two factors affecting the rate of evaporation. → Temperature and surface area (also humidity, wind).
- Why do the flat portions appear on a heating curve? → They are the change-of-state stages where heat is absorbed as latent heat at constant temperature.
- Which is greater for water — latent heat of fusion or of vaporisation? → Vaporisation ($2.26\times10^6$ vs $3.36\times10^5$ J kg⁻¹).
- Why does the climate near the sea remain moderate? → Water's high specific heat makes the sea warm/cool slowly, moderating coastal temperatures.
- Define one calorie. → Heat to raise 1 g of water through 1 °C ($=4.18$ J).
15. Everyday Applications of Specific & Latent Heat
- Cooking pots have thin metal bases (low heat capacity, heat up fast) but insulating handles (wood/plastic).
- Ice keeps drinks cold for long because melting absorbs a large latent heat of fusion.
- Cold drinks with ice cool more effectively than with cold water, since the ice must absorb its latent heat (336 kJ/kg) as it melts.
- Pressure cooker: higher pressure raises water's boiling point above 100 °C, cooking food faster.
- Land and sea breezes: land heats up faster by day (sea breeze blows inland) and cools faster by night (land breeze blows seaward) — a direct result of water's high specific heat.
16. Final Worked Numericals
Q11. 200 g of water at 30 °C is cooled by adding 20 g of ice at 0 °C. Find the final temperature (ignore the container). ($L_f=336$ J g⁻¹, $c=4.2$ J g⁻¹ °C⁻¹) Solution: heat to melt ice $=20\times336=6720$ J. Let final temp be T. Heat lost by water + melted ice warming: $200\times4.2\times(30-T) = 6720 + 20\times4.2\times(T-0)$. $840(30-T)=6720+84T \Rightarrow 25200-840T=6720+84T \Rightarrow 18480=924T \Rightarrow T=\mathbf{20\ °C}$.Q12. A copper calorimeter (heat capacity 60 J °C⁻¹) holds 100 g water at 20 °C. How much heat raises both to 50 °C? ($c_w=4.2$ J g⁻¹ °C⁻¹) Solution: water: $100\times4.2\times30=12600$ J; calorimeter: $60\times30=1800$ J; total $=\mathbf{14400\ J}$.
17. Final Quick-Revision Q&A
- Why are cooking-vessel handles made of wood or plastic? → They are poor conductors (insulators), so they stay cool to hold.
- Why does ice cool a drink better than cold water at 0 °C? → Ice additionally absorbs its latent heat of fusion (336 kJ/kg) as it melts.
- Why does food cook faster in a pressure cooker? → Increased pressure raises the boiling point of water above 100 °C.
- Why is the heat capacity of a body not the same as specific heat capacity? → Heat capacity is for the whole body ($mc$); specific heat capacity is per kg.