2026-03-21 · Quantity-of-Heat-Notes

QUANTITY OF HEAT

Grade XI Physics – Thermal Physics (Part 2) | NEB Syllabus (2080)

Thermal Physics

QUANTITY OF HEAT | Grade XI |- RICHESH SHARMA

1. Newton’s Law of Cooling

Statement: The rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small (≤ 20°C) and the nature of the radiating surface remains unchanged.

Mathematically:

− dQ/dt ∝ (T − T₀)

Since dQ = mc dT, ⇒ − mc (dT/dt) = k (T − T₀)

∴ dT/dt = −K (T − T₀)

where K = k/(mc) is a constant.

Solution of the Differential Equation

Integrating: ∫ dT/(T − T₀) = −K ∫ dt

⇒ ln(T − T₀) = −Kt + C

⇒ T − T₀ = e^{−Kt + C} = A e^−Kt

∴ T = T₀ + A e^−Kt

This shows that temperature decays exponentially toward T₀.

Figure 1: Exponential cooling curve (T vs t) — approaches T₀ asymptotically.

Verification (NEB Practical Focus)

Plot ln(T − T₀) vs t → should give a straight line with slope = −K.

ln(T − T₀) = −Kt + ln A

2. Specific Heat of Liquid by Method of Cooling

This experimental method uses **Newton’s Law of Cooling** to compare the cooling rates of two liquids.

Principle: For two bodies of equal mass, same surface area, same initial excess temperature (T − T₀), the ratio of their specific heat capacities is inversely proportional to the ratio of their rates of cooling.

From Newton’s Law: dT/dt = −K(T − T₀), and K = k/(mc)

At same (T − T₀), (dT/dt) ∝ 1/(mc)

If masses are equal: (dT/dt) ∝ 1/c

∴ c₁ / c₂ = (dT/dt)₂ / (dT/dt)₁

In practice, we use average rate of cooling over a temperature range:

c₁ / c₂ = [(θ₂ − θ₁)/t₂] / [(θ₂ − θ₁)/t₁] = t₁ / t₂

where t₁, t₂ = time taken by liquid 1 and 2 to cool through same Δθ.

Example (NEB 2078 Practical): Water and unknown liquid (equal mass) cool from 50°C to 40°C. Water takes 120 s, liquid takes 90 s. Find c_liquid. (c_w = 4200 J kg⁻¹ K⁻¹)

Solution:

c_w / c_l = t_l / t_w (∵ same Δθ, same mass)

⇒ 4200 / c_l = 90 / 120 = 3/4

⇒ c_l = 4200 × 4 / 3 = 5600 J kg⁻¹ K⁻¹

3. Specific Heat of Solid by Method of Mixture

This is the most common NEB practical experiment (see Grade XI Lab Manual).

Principle: Based on the Principle of Calorimetry — heat lost by hot solid = heat gained by cold water + calorimeter.

Let:
m = mass of solid,
c = specific heat of solid (unknown),
T₁ = initial temp of solid,
m_w, c_w, T₂ = mass, sp. heat, temp of water,
m_cal, c_cal = mass & sp. heat of calorimeter (or use water equivalent W),
T = final equilibrium temperature.

Heat lost by solid: Q₁ = m c (T₁ − T)

Heat gained by water + calorimeter:
Q₂ = m_w c_w (T − T₂) + m_cal c_cal (T − T₂)
or = (m_w c_w + W) (T − T₂), where W = water equivalent.

By calorimetry: Q₁ = Q₂

m c (T₁ − T) = (m_w c_w + W) (T − T₂)

⇒ c = [(m_w c_w + W)(T − T₂)] / [m (T₁ − T)]

Figure 2: Method of mixture — hot solid added to calorimeter.

4. Change of Phase and Latent Heat

Change of Phase: The transformation of a substance from one state (solid, liquid, gas) to another at constant temperature is called change of phase.

During phase change:
→ Temperature remains constant.
→ Heat supplied is used to overcome intermolecular forces (not to raise temperature).
→ This heat is called Latent Heat.

Latent Heat (L): Heat energy per unit mass required to change the phase of a substance without change in temperature.
Unit: J kg⁻¹

Types:

Type	Symbol	Definition	Example
Latent Heat of Fusion	L_f	Heat to melt 1 kg solid → liquid at melting point	Ice → Water at 0°C
Latent Heat of Vaporization	L_v	Heat to vaporize 1 kg liquid → vapour at boiling point	Water → Steam at 100°C

Total heat for phase change: Q = m L

Important Values (NEB Reference)

Substance	L_f (J kg⁻¹)	L_v (J kg⁻¹)
Water / Ice	3.36 × 10⁵	2.26 × 10⁶
Lead	2.5 × 10⁴	8.7 × 10⁵
Oxygen	1.4 × 10⁴	2.1 × 10⁵

_f) Liquid Boiling (L_v) Vapour 0°C 100°C

Figure 3: Heating curve showing plateaus (latent heat regions). Temperature constant during phase change.

5. Measurement of Specific Latent Heat

(a) Latent Heat of Fusion of Ice (by Mixture Method)

Procedure: Warm water in calorimeter → add dried ice at 0°C → find final temp.

Heat gained by ice:
= Heat to melt ice + heat to warm melted ice (water) from 0°C to T
= m_iL_f + m_ic_w(T − 0)

Heat lost by (water + calorimeter):
= (m_wc_w + W)(T₁ − T)

By calorimetry:

m_iL_f + m_ic_wT = (m_wc_w + W)(T₁ − T)

⇒ L_f = [(m_wc_w + W)(T₁ − T) − m_ic_wT] / m_i

Example: A calorimeter (W = 0.02 kg) contains 0.2 kg water at 30°C. 0.02 kg ice at 0°C is added. Final temp = 15°C. Find L_f.

Solution:

Heat lost = (0.2 × 4200 + 0.02 × 4200)(30 − 15) = (840 + 84) × 15 = 924 × 15 = 13,860 J

Heat gained by ice = 0.02 × L_f + 0.02 × 4200 × 15 = 0.02L_f + 1260

Equating: 0.02L_f + 1260 = 13860

⇒ 0.02L_f = 12600

⇒ L_f = 12600 / 0.02 = 6.3 × 10⁵ J kg⁻¹ (≈ theoretical 3.36×10⁵? → error due to assumption; real experiment accounts for heat loss)

Note: NEB expects correct formula application — value may differ due to idealization.

(b) Latent Heat of Vaporization of Water (by Condensation Method)

Procedure: Pass steam at 100°C into cold water in calorimeter. Collect condensed steam (mass m_s).

Heat lost by steam:
= Heat released during condensation + heat released by condensed water cooling from 100°C to T
= m_sL_v + m_sc_w(100 − T)

Heat gained by (water + calorimeter):
= (m_wc_w + W)(T − T₁)

By calorimetry:

m_sL_v + m_sc_w(100 − T) = (m_wc_w + W)(T − T₁)

⇒ L_v = [(m_wc_w + W)(T − T₁) − m_sc_w(100 − T)] / m_s

Example (NEB 2076): 0.01 kg steam at 100°C is passed into 0.24 kg water at 20°C in a 0.02 kg copper calorimeter. Final temp = 40°C. Find L_v. (c_w=4200, c_Cu=385 J kg⁻¹ K⁻¹)

Solution:

W = m_calc_cal/c_w = (0.02 × 385)/4200 = 0.00183 kg (or use directly)

Heat gained = (0.24×4200 + 0.02×385)(40−20) = (1008 + 7.7) × 20 = 1015.7 × 20 = 20,314 J

Heat lost by steam = 0.01 L_v + 0.01×4200×(100−40) = 0.01L_v + 2520

Equating: 0.01L_v + 2520 = 20314

⇒ 0.01L_v = 17794

⇒ L_v = 1.7794 × 10⁶ J kg⁻¹ ≈ 1.78 × 10⁶ J kg⁻¹ (close to 2.26×10⁶ — error due to heat loss; NEB accepts method)

6. NEB-Style MCQs

Choose the correct answer (1 mark each)

Q1. Newton’s law of cooling is valid when:
(a) Temperature difference > 30°C (b) Body is black
(c) Temperature difference is small (≤20°C) (d) Convection dominates

Q2. In the method of mixture, the calorimeter’s heat capacity is often expressed as:
(a) Specific heat (b) Water equivalent (c) Thermal conductivity (d) Emissivity

Q3. During melting of ice, the heat supplied is used to:
(a) Increase kinetic energy (b) Increase temperature
(c) Break molecular bonds (d) Increase pressure

Q4. The SI unit of latent heat is:
(a) J (b) J kg⁻¹ (c) J K⁻¹ (d) cal g⁻¹ °C⁻¹

Q5. When steam condenses to water at 100°C, it:
(a) Absorbs latent heat (b) Releases latent heat
(c) Temperature increases (d) Density decreases

Answer Key:

Q1: (c) Temperature difference is small (≤20°C)
Q2: (b) Water equivalent
Q3: (c) Break molecular bonds
Q4: (b) J kg⁻¹
Q5: (b) Releases latent heat

7. NEB Exam Tips

🔹 Newton’s Law of Cooling: Must know derivation (1st order DE), graph, and verification method (ln(T−T₀) vs t).
🔹 Method of Cooling (liquid): c ∝ 1/t (equal mass, same Δθ). NEB often asks 2-mark numerical.
🔹 Method of Mixture (solid): Most important practical — memorize formula for c.
🔹 Latent Heat Experiments: Focus on energy balance equations (L_f and L_v). NEB gives 4-mark numerical.
🔹 Always write assumptions: no heat loss, complete mixing, steam/dry ice at exact phase-change temp.
🔹 Use water equivalent (W) when calorimeter material is given.