Unit 11: Quantity of Heat

Introduction

Heat is a form of energy that flows from a body at higher temperature to one at lower temperature. Understanding how heat is transferred, stored, and released during temperature changes and phase transitions is fundamental to many applications in physics, engineering, and everyday life. This chapter provides a detailed treatment of the quantitative aspects of heat, including cooling laws, calorimetry, latent heat, and the unique triple point condition.

Newton's Law of Cooling

Definition and Mathematical Form

Newton's Law of Cooling states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small and the nature of the surface remains unchanged.

The law can be expressed as:

dQ/dt = -k (T - T_s)

where:

dQ/dt is the rate of heat loss (J/s or W),
k is the positive constant of proportionality (depends on surface area, nature of surface, and surrounding medium),
T is the instantaneous temperature of the body (K or °C),
T_s is the temperature of the surroundings (K or °C).

The negative sign indicates that heat flows out of the body when T > T_s. Integrating this differential equation yields an exponential decay of temperature with time:

T(t) = T_s + (T_0 - T_s) e^{-kt}

where T_0 is the initial temperature of the body at t = 0. Consequently, a plot of temperature versus time shows a curve that approaches the surrounding temperature asymptotically.

Applications

Newton's Law of Cooling finds practical use in:

Estimating the time of death in forensic science (cooling of a corpse),
Determining the cooling time of hot beverages or molten metals,
Designing heat exchangers and thermal management systems.

Numerical Problem 1

A hot metal sphere at 80 °C is placed in a room where the ambient temperature is 20 °C. After 5 minutes, its temperature drops to 50 °C. Assuming Newton's Law of Cooling applies, determine the constant k and predict the temperature after an additional 10 minutes.

Solution:

Use the integrated form: T(t) = T_s + (T_0 - T_s) e^{-kt}.
Insert known values for t = 5 min (=300 s): 50 = 20 + (80 - 20) e^{-k·300} → 30 = 60 e^{-300k} → e^{-300k} = 0.5.
Take natural log: -300k = ln(0.5) = -0.6931 → k = 0.6931/300 ≈ 2.31×10^{-3} s^{-1}.
Now find temperature after total t = 15 min (=900 s): T(900) = 20 + 60 e^{- (2.31×10^{-3})(900)}.
Compute exponent: -2.31×10^{-3} × 900 = -2.079 → e^{-2.079} ≈ 0.125.
Thus T(900) = 20 + 60 × 0.125 = 20 + 7.5 = 27.5 °C.

After an additional 10 minutes (total 15 minutes), the sphere’s temperature will be approximately 27.5 °C.

Measurement of Specific Heat Capacity

Specific heat capacity (c) is the amount of heat required to raise the temperature of unit mass of a substance by one kelvin (or one degree Celsius). Its SI unit is joule per kilogram per kelvin (J kg⁻¹ K⁻¹); in the CGS system it is cal g⁻¹ °C⁻¹.

Solids: Method of Mixtures (Calorimetry)

When a hot solid is placed in a calorimeter containing a known mass of water (or another liquid) at a lower temperature, heat lost by the hot body equals heat gained by the calorimeter and the liquid, assuming no heat loss to the surroundings.

The heat balance equation is:

m₁ c₁ (T₁ - T_f) = m₂ c₂ (T_f - T₂) + m_c c_c (T_f - T_c)

where:

m₁, c₁ – mass and specific heat of the hot solid,
T₁ – initial temperature of the solid,
m₂, c₂ – mass and specific heat of the liquid (usually water, c₂ ≈ 4186 J kg⁻¹ K⁻¹),
T₂ – initial temperature of the liquid,
m_c, c_c – mass and specific heat of the calorimeter (often made of copper or aluminium),
T_f – final equilibrium temperature of the mixture.

If the calorimeter’s heat capacity is known (C_c = m_c c_c), the equation simplifies to:

m₁ c₁ (T₁ - T_f) = m₂ c₂ (T_f - T₂) + C_c (T_f - T_c)

Numerical Problem 2 (Solid)

A 0.250 kg block of an unknown metal is heated to 120 °C and then quickly transferred into a calorimeter containing 0.150 kg of water at 22 °C. The calorimeter (made of copper) has a mass of 0.050 kg and a specific heat of 385 J kg⁻¹ K⁻¹. The final temperature of the mixture is 28 °C. Calculate the specific heat capacity of the metal.

Solution:

Heat lost by metal: Q_loss = m_m c_m (T_m - T_f) = 0.250·c_m·(120 - 28) = 0.250·c_m·92.
Heat gained by water: Q_w = m_w c_w (T_f - T_w) = 0.150·4186·(28 - 22) = 0.150·4186·6.
Heat gained by calorimeter: Q_c = m_c c_c (T_f - T_c) = 0.050·385·(28 - 22) = 0.050·385·6.
Set heat lost = heat gained: 0.250·c_m·92 = 0.150·4186·6 + 0.050·385·6.
Compute RHS: 0.150·4186·6 = 3767.4 J; 0.050·385·6 = 115.5 J; sum = 3882.9 J.
Thus 0.250·92·c_m = 3882.9 → 23.0·c_m = 3882.9 → c_m = 3882.9 / 23.0 ≈ 168.8 J kg⁻¹ K⁻¹.

The specific heat capacity of the metal is approximately 169 J kg⁻¹ K⁻¹.

Liquids: Electrical Method

In the electrical method, a known electric energy is supplied to a liquid via a heating coil immersed in it. The temperature rise of the liquid is measured, and the specific heat is calculated from the energy balance.

The electrical energy supplied is:

E = V I t

where V is voltage (volts), I is current (amperes), and t is time (seconds). Assuming negligible heat losses, this energy equals the heat absorbed by the liquid:

E = m c ΔT

Hence:

c = \frac{V I t}{m ΔT}

Numerical Problem 3 (Liquid – Electrical)

0.350 kg of an organic liquid is placed in an insulated container. A heating coil supplied with a constant voltage of 12 V and a current of 2.5 A is immersed in the liquid for 8 minutes. The temperature of the liquid rises from 18 °C to 34 °C. Determine the specific heat capacity of the liquid.

Solution:

Electrical energy: E = V I t = 12·2.5·(8×60) = 12·2.5·480 = 14400 J.
Mass m = 0.350 kg, temperature change ΔT = 34 - 18 = 16 °C.
Specific heat: c = E / (m ΔT) = 14400 / (0.350·16) = 14400 / 5.6 ≈ 2571.4 J kg⁻¹ K⁻¹.

The liquid’s specific heat capacity is about 2.57 kJ kg⁻¹ K⁻¹.

Liquids: Continuous Flow Method

In this method, the liquid flows at a steady rate through a heated section where a known power is supplied. By measuring the inlet and outlet temperatures and the flow rate, the specific heat can be obtained from:

P = \dot{m} c (T_{out} - T_{in})

where \dot{m} is the mass flow rate (kg s⁻¹) and P is the heating power (W). Rearranged:

c = \frac{P}{\dot{m} (T_{out} - T_{in})}

This technique is advantageous for studying liquids that may undergo chemical change if heated for long periods.

Units of Specific Heat Capacity

Common units include:

SI: joule per kilogram per kelvin (J kg⁻¹ K⁻¹),
CGS: calorie per gram per degree Celsius (cal g⁻¹ °C⁻¹) (1 cal = 4.184 J).

Typical values: water ≈ 4186 J kg⁻¹ K⁻¹, aluminium ≈ 900 J kg⁻¹ K⁻¹, iron ≈ 450 J kg⁻¹ K⁻¹.

Change of Phases: Latent Heat

During a phase transition (solid↔liquid, liquid↔gas, solid↔gas) the temperature of the substance remains constant despite continuous heat exchange. The energy supplied or removed is used to overcome intermolecular forces rather than increase kinetic energy. This heat is termed latent heat.

Latent heat (L) is defined as the amount of heat required per unit mass to effect a phase change at constant temperature and pressure.

Q = m L

where Q is the heat exchanged, m is the mass undergoing the phase change, and L is the specific latent heat (J kg⁻¹).

Numerical Problem 4 (Latent Heat – Melting Ice)

How much heat is required to melt 0.750 kg of ice at 0 °C? Use the latent heat of fusion of ice, L_f = 334 kJ kg⁻¹.

Solution:

Q = m L_f = 0.750·334 000 = 250 500 J ≈ 2.51×10⁵ J.

Thus 251 kJ of heat must be supplied.

Specific Latent Heat

Specific latent heat distinguishes the two most common phase changes:

Latent heat of fusion (L_f): heat needed to convert 1 kg of solid into liquid at its melting point.
Latent heat of vaporization (L_v): heat needed to convert 1 kg of liquid into vapor at its boiling point.

Representative values for water:

Property	Symbol	Value
Latent heat of fusion	`L_f`	334 kJ kg⁻¹
Latent heat of vaporization	`L_v`	2260 kJ kg⁻¹

Numerical Problem 5 (Vaporization)

Calculate the energy required to vaporize 0.120 kg of water at 100 °C.

Solution:

Q = m L_v = 0.120·2260 000 = 271 200 J ≈ 2.71×10⁵ J.

Thus about 271 kJ of energy is needed.

Measurement of Specific Latent Heat

Fusion: Method of Mixtures with Ice

To determine L_f of a substance (commonly ice), a known mass of ice at 0 °C is added to a known mass of warm water in a calorimeter. The ice melts, absorbing heat, while the warm water cools down. Assuming no heat loss:

m_w c_w (T_w - T_f) = m_i L_f + m_i c_w (T_f - 0)

where:

m_w, c_w – mass and specific heat of the warm water,

T_w – initial temperature of warm water,

T_f – final temperature after all ice has melted (should be >0 °C if all ice melts),

m_i – mass of ice,

L_f – latent heat of fusion (unknown),

The term m_i c_w (T_f - 0) accounts for heating the melted ice (now water) from 0 °C to the final temperature.

Re‑arranging gives:

L_f = \frac{m_w c_w (T_w - T_f) - m_i c_w T_f}{m_i}

Numerical Problem 6 (Fusion Measurement)

In an experiment, 0.080 kg of ice at 0 °C is added to 0.250 kg of water initially at 45 °C in a calorimeter of negligible heat capacity. After equilibrium, the final temperature is 12 °C. Determine the experimental value of L_f for ice.

Solution:

Heat lost by warm water: Q_lost = m_w c_w (T_w - T_f) = 0.250·4186·(45 - 12) = 0.250·4186·33.

Compute: 0.250·4186 = 1046.5; 1046.5·33 = 34 534.5 J.

Heat gained by ice consists of two parts: melting (m_i L_f) and warming the resulting water (m_i c_w T_f).

Heat gained by warming melted ice: Q_warm = m_i c_w T_f = 0.080·4186·12 = 0.080·4186·12.

Compute: 0.080·4186 = 334.88; 334.88·12 = 4 018.56 J.

Set heat lost = heat gained: 34 534.5 = m_i L_f + 4 018.56.

Thus m_i L_f = 34 534.5 - 4 018.56 = 30 515.94 J.

Finally, L_f = 30 515.94 / m_i = 30 515.94 / 0.080 = 381 449.25 J kg⁻¹ ≈ 381 kJ kg⁻¹.

The obtained value (~381 kJ kg⁻¹) is higher than the accepted 334 kJ kg⁻¹, indicating probable heat losses or experimental error.

Vaporization: Steam Injection Method

The specific latent heat of vaporization of water can be measured by injecting a known mass of dry steam at 100 °C into a known mass of water at a lower temperature. The steam condenses, releasing its latent heat, and the resulting water (from condensed steam) mixes with the original water, raising the temperature.

Heat balance (assuming negligible losses and that the calorimeter’s heat capacity is known or included):

m_s L_v + m_s c_w (T_f - 100) = m_w c_w (T_f - T_w) + C_c (T_f - T_c)

where:

m_s – mass of steam injected,

L_v – latent heat of vaporization (unknown),

m_s c_w (T_f - 100) – heat released when the condensed steam (now water at 100 °C) cools to final temperature T_f (note the term is negative if T_f < 100; we keep it on left side as heat released),

m_w, c_w, T_w – mass, specific heat, and initial temperature of the original water,

C_c, T_c – heat capacity and initial temperature of the calorimeter (often taken equal to T_w if initially in equilibrium).

If the calorimeter’s heat capacity is negligible or already accounted for, the equation simplifies to:

L_v = \frac{m_w c_w (T_f - T_w) - m_s c_w (T_f - 100)}{m_s}

Numerical Problem 7 (Vaporization Measurement)

0.020 kg of steam at 100 °C is injected into 0.300 kg of water initially at 20 °C in a calorimeter of negligible heat capacity. After condensation and mixing, the final temperature is 38 °C. Determine the experimental latent heat of vaporization of water.

Solution:

Heat gained by original water: Q_w = m_w c_w (T_f - T_w) = 0.300·4186·(38 - 20) = 0.300·4186·18.

Compute: 0.300·4186 = 1255.8; 1255.8·18 = 22 604.4 J.

Heat released by steam consists of latent heat (m_s L_v) plus cooling of the condensed water from 100 °C to T_f.

Cooling term: Q_cool = m_s c_w (100 - T_f) = 0.020·4186·(100 - 38) = 0.020·4186·62 (note we treat this as heat released, so it adds to the left side).

Compute: 0.020·4186 = 83.72; 83.72·62 = 5 190.64 J.

Total heat released by steam: Q_released = m_s L_v + Q_cool.

Set heat gained = heat released: 22 604.4 = m_s L_v + 5 190.64.

Thus m_s L_v = 22 604.4 - 5 190.64 = 17 413.76 J.

Finally, L_v = 17 413.76 / m_s = 17 413.76 / 0.020 = 870 688 J kg⁻¹ ≈ 871 kJ kg⁻¹.

The obtained value (~871 kJ kg⁻¹) is lower than the accepted 2260 kJ kg⁻¹, indicating significant heat losses or incomplete condensation in this simplified example.

Triple Point

The triple point of a substance is the unique combination of temperature and pressure at which the solid, liquid, and gas phases coexist in thermodynamic equilibrium. At this point, the substance can change between any two phases without a net change in the other phase’s amount.

For water, the triple point occurs at:

Temperature: T_t = 273.16 K (which is exactly 0.01 °C),
Pressure: P_t = 611.657 Pa (approximately 6.12 mbar).
This point is used to define the Kelvin scale: the kelvin is defined as the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. On a phase diagram (pressure vs. temperature), the triple point appears as the intersection of the three phase‑boundary lines (solid‑liquid, liquid‑gas, solid‑gas). It is a singular point; moving away from it in any direction causes one of the phases to disappear. Numerical Problem 8 (Triple Point Concept) If a sample of water is placed in a sealed container and the pressure is reduced to 600 Pa while the temperature is held at 273.15 K, which phases will be present? Solution: At 273.15 K (0.00 °C) the temperature is just below the triple‑point temperature (273.16 K). The pressure of 600 Pa is slightly below the triple‑point pressure (≈612 Pa). On the phase diagram, this condition lies in the region where solid and vapor coexist (the sublimation curve). Therefore, the container will contain solid ice and water vapor, with no liquid phase present. Summary This chapter has presented the quantitative framework for heat transfer: Newton's Law of Cooling describes exponential temperature decay and enables practical estimations such as cooling times and forensic applications. Specific heat capacity can be measured via calorimetry (method of mixtures) for solids and via electrical or continuous‑flow methods for liquids. During phase changes, temperature remains constant; the associated heat is latent heat, quantified by specific latent heats of fusion and vaporization. Experimental determination of these latent heats relies on heat‑balance principles—melting ice with warm water and condensing steam into cooler water. The triple point of water (273.16 K, 612 Pa) uniquely defines the coexistence of solid, liquid, and gas and underpins the definition of the kelvin. Mastery of these concepts provides the foundation for further studies in thermodynamics, heat engineering, and environmental science.