5. Heat​

Chapter 5 — Heat

Latent Heat Regelation Anomalous Behaviour of Water Dew Point & Humidity Specific Heat Capacity Heat Exchange & Calorimetry

Can you recall?

  • Heat vs Temperature: Heat is energy in transit; temperature measures hotness and is proportional to average kinetic energy.
  • Modes of transfer: Conduction, convection, radiation.
  • Regelation: Melting of ice under pressure and refreezing when pressure is released.
  • Dew point & Humidity: Measures of moisture in air and the temperature at which condensation starts.

Latent Heat

During a change of state, a substance absorbs or releases heat at a constant temperature. This hidden heat is called latent heat.

\[ \textbf{Fusion:}\quad Q = m\,L_f \qquad\qquad \textbf{Vaporization:}\quad Q = m\,L_v \]
  • Melting point: Constant temperature at which solid → liquid (for ice, \(0^\circ\text{C}\) at 1 atm).
  • Boiling point: Constant temperature at which liquid → gas (for water, \(100^\circ\text{C}\) at 1 atm).
  • Specific Latent Heat of Fusion \(L_f\): Heat absorbed by unit mass of a solid to become liquid at its melting point.
  • Specific Latent Heat of Vaporization \(L_v\): Heat absorbed by unit mass of a liquid to become gas at its boiling point.
For water at 1 atm: \(L_f \approx 333~\text{kJ kg}^{-1}\) \((80~\text{cal g}^{-1})\) and \(L_v \approx 2256~\text{kJ kg}^{-1}\) \((540~\text{cal g}^{-1})\).

Heating Curve (Ice → Water → Steam)

As you heat ice, temperature stays at \(0^\circ\text{C}\) while melting (segment AB), then rises to \(100^\circ\text{C}\) (segment BC), and stays at \(100^\circ\text{C}\) during boiling (segment CD).

A B C D Temperature Time Melting @ 0°C Boiling @ 100°C
Pressure matters: Melting/boiling points and latent heats depend on external pressure.

Regelation

Regelation is the phenomenon where ice melts under pressure and refreezes when the pressure is removed. It explains how an ice-ball forms and why a taut wire with weights passes through an ice block without splitting it.

  • Pressure lowers the melting point of ice: under pressure, part of the ice becomes water at \(0^\circ\text{C}\).
  • When pressure is released, melting point returns to \(0^\circ\text{C}\); the water refreezes, “healing” behind the wire.

Anomalous Behaviour of Water

Unlike most liquids, water contracts when heated from \(0^\circ\text{C}\) to \(4^\circ\text{C}\), reaching maximum density at \(4^\circ\text{C}\); beyond \(4^\circ\text{C}\), it expands normally.

Hope’s Apparatus — what happens?

  • Cooling from the middle makes denser (\(\approx4^\circ\text{C}\)) water sink; lower thermometer \(T_1\) rapidly approaches \(4^\circ\text{C}\) and lingers.
  • Water cooled below \(4^\circ\text{C}\) becomes less dense and rises; upper thermometer \(T_2\) reaches \(0^\circ\text{C}\) earlier.
Life saver for aquatic organisms: In cold regions, surface water freezes first; deeper water stays near \(4^\circ\text{C}\), allowing plants and animals to survive under ice.
Frost weathering: Water in pipes/rocks expands on freezing → bursting/cracking in winters.

Dew Point and Humidity

Humidity is moisture in air. If the air contains the maximum vapour possible at that temperature, it is saturated. Cooling unsaturated air increases its relative humidity until saturation at the dew point temperature, when condensation starts.

Absolute & Relative Humidity

  • Absolute humidity (AH): mass of water vapour per unit volume of air (e.g., kg m\(^{-3}\)).
  • Relative humidity (RH):
    \[ \%\,\text{RH} \;=\; \dfrac{\text{actual vapour mass (at T) in a given volume}}{\text{vapour mass needed for saturation (at T)}}\times 100 \]
Observe: Water droplets on a cold bottle/early morning leaves appear when surrounding air is cooled to (or below) its dew point.

Unit of Heat

  • SI unit: Joule (J).
  • cgs unit: calorie (cal) — heat to raise \(1~\text{g}\) water by \(1^\circ\text{C}\) (from \(14.5^\circ\) to \(15.5^\circ\text{C}\)).
\[ 1~\text{cal} \approx 4.18~\text{J}, \qquad 1~\text{kcal} = 1000~\text{cal} \]
James P. Joule: Demonstrated mechanical equivalence of heat; the heat unit “Joule” is named after him.

Specific Heat Capacity

The specific heat capacity \(c\) of a substance is the heat required to raise the temperature of unit mass by \(1^\circ\text{C}\).

\[ Q = m\,c\,\Delta T \qquad\text{(heat absorbed if }\Delta T>0\text{; heat lost if }\Delta T<0\text{)} \]

Units: SI: J kg\(^{-1}\) °C\(^{-1}\); cgs: cal g\(^{-1}\) °C\(^{-1}\).

Why do equal masses heat differently?

Metals/s liquids have different \(c\). In the classic “wax slab” test (iron, copper, lead spheres heated to \(100^\circ\text{C}\)), the sphere giving more heat melts more wax and sinks deeper — showing differing specific heats.

Typical Values (at ~room temperature)

SubstanceSpecific heat (cal g\(^{-1}\) °C\(^{-1}\))SubstanceSpecific heat (cal g\(^{-1}\) °C\(^{-1}\))
Water1.000Aluminium0.215
Paraffin0.540Copper0.095
Kerosene0.520Silver0.056
Iron0.110Mercury0.033

Heat Exchange & Calorimetry

When a hot object is mixed with a cold object in an insulated system, heat lost by the hot one equals heat gained by the cold one until common final temperature \(T_f\) is reached.

\[ \text{Isolated system:}\quad Q_{\text{lost}} = Q_{\text{gained}} \] \[ m_h c_h\,(T_h - T_f)\;=\; m_c c_c\,(T_f - T_c)\;+\; m_{\text{cal}} c_{\text{cal}}\,(T_f - T_c) \]

In a calorimeter, a hot solid is placed in water. Measure initial temperatures, then the common final temperature. Using known \(c\) for water & calorimeter, compute the unknown \(c\) of the solid from the heat balance.

Mixing method: Carefully stir; minimize heat loss; use a heat-resistant, well-lagged calorimeter to approximate an isolated system.

Think • Try • Apply

  • Where does latent heat go? Into breaking/forming intermolecular bonds during phase change (not into raising temperature).
  • Regelation & latent heat: Local melting absorbs \(mL_f\); refreezing releases the same amount when pressure is removed.
  • Humidity feels: RH \(>\) 60% → humid; RH \(<\) 60% → dry. At dew point, RH = \(100\%\).
  • Water’s anomaly: Explains survival of aquatic life under ice and bursting of water pipes in winter.
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Chapter 5 — Heat: Exercise Solutions

Latent Heat Specific Heat Capacity Regelation & Anomaly of Water Humidity & Dew Point Heat Exchange

1) Fill in the blanks

  • a) The amount of water vapour in air is determined in terms of its humidity.
  • b) If equal masses receive equal heat but reach different final temperatures, it is due to difference in their specific heat capacities.
  • c) During transformation of liquid phase to solid phase, the latent heat is released (given out).

2) Water’s volume–temperature graph (from \(0^\circ\)C upwards)

Most substances expand on heating: their volume rises monotonically with temperature. Water is exceptional between \(0^\circ\text{C}\) and \(4^\circ\text{C}\): its volume decreases as it is heated from \(0^\circ\)C to \(4^\circ\)C (so density increases), then above \(4^\circ\)C it expands normally.

Name of behaviour: Anomalous behaviour of water (maximum density at \(4^\circ\)C).

3) Specific heat capacity & experiment

Definition: The specific heat capacity \(c\) of a substance is the heat needed to raise temperature of unit mass by \(1^\circ\text{C}\).

\(Q = m\,c\,\Delta T\)

Proof that different substances have different \(c\)

Wax-slab test: Heat equal-mass spheres (iron, copper, lead) to the same temperature (e.g., \(100^\circ\)C), then place on a thick wax slab. The sphere that melts more wax (sinks deeper) gives out more heat, showing a larger product \(m c \Delta T\). With same \(m\) and \(\Delta T\), the differences arise from different \(c\) values.
Observation typically: Iron sinks most > Copper > Lead ⇒ \(c_{\text{Fe}} > c_{\text{Cu}} > c_{\text{Pb}}\).

4) Temperature interval chosen for the unit of heat (calorie)

By definition, 1 calorie is the heat required to raise \(1~\text{g}\) of water from \(14.5^\circ\)C to \(15.5^\circ\)C. This narrow interval near room temperature is chosen because water’s heat capacity varies slightly with temperature; fixing a standard interval ensures reproducible measurement.

5) Explain the temperature–time graph (ice → water → steam)

  • Segment (Ice + Water at \(0^\circ\)C): Temperature remains constant at \(0^\circ\)C while ice melts. Heat supplied is latent heat of fusion—used to break bonds, not raise temperature.
  • Rising Water Region: After all ice melts, water warms from \(0^\circ\)C to \(100^\circ\)C; temperature rises with time.
  • Plateau at \(100^\circ\)C (Boiling water + vapour): Temperature stays at \(100^\circ\)C while water changes to steam. Heat supplied is latent heat of vaporization.

6) Explanations

a) Aquatic life in cold regions: Because water is densest at \(4^\circ\)C, the bottom of lakes stays near \(4^\circ\)C even when the surface freezes. The ice layer insulates, allowing organisms to survive below.

b) Dew vs droplets on a cold bottle: Air near a cold surface cools to its dew point; excess water vapour condenses as tiny drops. A chilled bottle mimics plants/grass at dawn—both show condensation when air hits its dew point.

c) Rock cracking in winter: Water in cracks freezes and expands (anomalous expansion), exerting pressure that widens cracks—eventually causing rock fracture (frost weathering).

7) Short answers

a) Latent heat & state change when given off: Latent heat is the heat absorbed/released during a change of state at constant temperature. When latent heat is released, matter changes to a more ordered phase: gas \(\to\) liquid (condensation) or liquid \(\to\) solid (freezing).

b) Principle used to measure \(c\): Principle of Heat Exchange (in an isolated system, heat lost by hot body equals heat gained by cold body) via calorimetry.

c) Role of latent heat in state change: Supplied latent heat breaks intermolecular bonds (solid \(\to\) liquid, liquid \(\to\) gas) without raising temperature; released latent heat forms bonds during condensation/freezing, again at constant temperature.

d) How to tell if air is saturated: Air is saturated when \(\text{RH}=100\%\) (actual vapour = saturation vapour at that \(T\)). Determine by measuring temperature & humidity (hygrometer/psychrometer) or by dew point—if the air’s temperature equals its dew point, it is saturated.

8) Based on the paragraph

  • i) Heat transfers from the hot object to the cold object.
  • ii) Principle of Heat Exchange / Energy conservation in thermal interactions.
  • iii) In an isolated system: \(\boxed{\,\text{Heat lost by hot body} = \text{Heat gained by cold body}\,}\).
  • iv) It enables measurement of specific heat capacity (and also latent heats) via calorimetry.

9) Numerical Problems (with perfect working)

(a) Equal heat to A and B (1 g each) raises \(T_A\) by \(3^\circ\)C, \(T_B\) by \(5^\circ\)C

\[ Q = m\,c\,\Delta T \;\Rightarrow\; c \propto \frac{Q}{m\,\Delta T}\;\;(\text{given }Q,m\text{ same})\;\Rightarrow\; \frac{c_A}{c_B}=\frac{\Delta T_B}{\Delta T_A}=\frac{5}{3} \]
Answer: Object A has more specific heat; factor \(=\; \dfrac{5}{3}\).

(b) Ammonia needed to make 2 kg ice from water at \(20^\circ\)C

\[ Q_{\text{remove}} = m_w c_w \Delta T + m_w L_f = 2000(1)(20) + 2000(80) = 200{,}000\ \text{cal} \] \[ m_{\text{NH}_3}=\frac{Q_{\text{remove}}}{L_v(\text{NH}_3)}=\frac{200{,}000}{341}\approx 586.4\ \text{g} \]
Answer: \(\approx \mathbf{586.4~g}\) of ammonia.

(c) \(150\) g ice at \(0^\circ\)C mixed with steam at \(100^\circ\)C to get water at \(50^\circ\)C

Let \(m_s\) = mass of steam (g). Steam \(\to\) water at \(100^\circ\)C (releases \(540\) cal/g), then cools to \(50^\circ\)C (releases \(50\) cal/g): total \(590\) cal/g.

Ice: melt \(80\) cal/g, then warm to \(50^\circ\)C: \(50\) cal/g ⇒ total \(130\) cal/g. For \(150\) g: \(150\times130=19{,}500\) cal.

\[ 590\,m_s = 19{,}500 \;\Rightarrow\; m_s \approx 33.05\ \text{g} \]
Answer: \(\approx \mathbf{33~g}\) steam.

(d) Calorimeter problem (final temperature)

Data: \(m_{\text{cal}}=100\) g, \(c_{\text{cal}}=0.1\ \text{cal g}^{-1}{}^\circ\text{C}^{-1}\); liquid: \(250\) g, \(c=0.4\ \text{cal g}^{-1}{}^\circ\text{C}^{-1}\), initial \(30^\circ\)C; ice: \(10\) g at \(0^\circ\)C.

Assume all ice melts and final \(T_f>0^\circ\)C. Heat lost by (liquid + calorimeter) equals heat gained by (ice melting + melted water warming).

\[ (250\cdot0.4 + 100\cdot0.1)(30 - T_f) = 10\cdot80 + 10\cdot( T_f - 0) \] \[ (100+10)(30-T_f) = 800 + 10T_f \;\Rightarrow\; 110(30-T_f)=800+10T_f \] \[ 3300 - 110T_f = 800 + 10T_f \;\Rightarrow\; 2500 = 120T_f \;\Rightarrow\; T_f \approx 20.8^\circ\text{C} \]
Answer: Final temperature \(\approx \mathbf{20.8^\circ\textbf{C}}\).

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