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Temperature, Heat & Calorimetry

Master Temperature, Heat and Calorimetry for JEE Main, JEE Advanced, NEET & Board Exams. Covers temperature scales, Zeroth Law, specific heat, heat capacity, principle of calorimetry, latent heat, heating curve of water, water equivalent and Joule's mechanical equivalent of heat with solved examples.

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Thermal Physics begins with the most basic question: what is heat and how do we measure it? Temperature is the degree of hotness or coldness of a body, while heat is the energy that flows between bodies due to a temperature difference. Calorimetry is the science of measuring heat exchange in physical and chemical processes. These foundational concepts underpin all of thermodynamics and are directly tested in JEE and NEET — with questions on temperature scales, specific heat, calorimetry problems, and latent heat appearing every year. For board exams, this chapter is essential for both theory and numericals.

1. Temperature and Its Measurement

Temperature is a measure of the average kinetic energy of the molecules of a substance. It determines the direction of heat flow — heat always flows from higher temperature to lower temperature, until thermal equilibrium is reached.

Zeroth Law of Thermodynamics

If body A is in thermal equilibrium with body C, and body B is also in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other. This law forms the basis of temperature measurement using a thermometer — the thermometer (C) is the reference body.

Temperature Scales

Scale	Ice Point	Steam Point	Absolute Zero	Symbol
Celsius	$0 °$ C	$100 °$ C	$- 273.15 °$ C	°C
Fahrenheit	$32 °$ F	$212 °$ F	$- 459.67 °$ F	°F
Kelvin (SI)	$273.15$ K	$373.15$ K	$0$ K	K
Rankine	$491.67 °$ R	$671.67 °$ R	$0 °$ R	°R

Temperature Conversion Formulae

$\frac{C - 0}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}$

The general relation between any two scales X and Y with ice points $X_{0}$ , $Y_{0}$ and steam points $X_{100}$ , $Y_{100}$ :

$\frac{X - X_{0}}{X_{100} - X_{0}} = \frac{Y - Y_{0}}{Y_{100} - Y_{0}}$

Key Conversion Results

$T (K) = T (°C) + 273.15 \approx T (°C) + 273$
$T (°F) = \frac{9}{5} T (°C) + 32$
$T (°C) = \frac{5}{9} (T (°F) - 32)$
Celsius and Fahrenheit are equal at $- 40 °$ : $C = F = - 40 °$
A change of $1 °$ C = a change of $\frac{9}{5} °$ F = a change of $1$ K

2. Ideal Gas Temperature Scale and Absolute Zero

The Kelvin (absolute) scale is the most fundamental temperature scale. It is based on the behaviour of an ideal gas — at constant volume, the pressure of an ideal gas is proportional to its absolute temperature:

$P \propto T$ (at constant $V$ and $n$ )

Absolute Zero (0 K = $- 273.15 °$ C) is the temperature at which an ideal gas would have zero pressure and zero volume — the molecules would have zero kinetic energy. It is the lowest theoretically possible temperature. In practice, absolute zero can be approached but never reached (Third Law of Thermodynamics).

At absolute zero, molecular motion does not completely cease due to quantum zero-point energy, but classical kinetic energy is zero.
Negative temperatures on the Kelvin scale are physically meaningless for equilibrium systems.

3. Heat and Thermal Energy

Heat (Q) is energy in transit — it flows from a body at higher temperature to a body at lower temperature due to the temperature difference. It is NOT a property stored in a body; it is a process quantity.

SI unit of heat: Joule (J). Also used: calorie (cal), kilocalorie (kcal), British Thermal Unit (BTU).
$1 cal = 4.186 J$ (mechanical equivalent of heat — established by Joule)
$1 kcal = 4186 J = 4.186 kJ$

Distinction: Heat vs Temperature vs Internal Energy

Concept	Definition	Nature
Temperature	Measure of average KE of molecules	State variable (property of body)
Heat	Energy transferred due to temperature difference	Process variable (not stored in body)
Internal Energy	Total KE + PE of all molecules of a body	State variable (stored in body)

4. Specific Heat Capacity

The specific heat capacity (or specific heat) of a substance is the amount of heat required to raise the temperature of unit mass by unit temperature:

$c = \frac{Q}{m Δ T}$ $\Rightarrow$ $Q = m c Δ T$

SI unit: J·kg⁻¹·K⁻¹ (also written J/kg/°C). Also commonly used: J/g/°C or cal/g/°C.

Molar Heat Capacity

The heat required to raise the temperature of one mole by one degree:

$C = \frac{Q}{n Δ T} = M c$

where $n$ is number of moles and $M$ is molar mass. SI unit: J·mol⁻¹·K⁻¹.

Specific Heat at Constant Pressure and Constant Volume

$c_{p}$ — specific heat at constant pressure (includes work done against external pressure): always larger than $c_{v}$ .
$c_{v}$ — specific heat at constant volume (all heat goes into internal energy): always smaller than $c_{p}$ .
For an ideal gas: $c_{p} - c_{v} = R / M$ (where $R = 8.314$ J/mol/K and $M$ is molar mass).
Molar: $C_{p} - C_{v} = R$ (Mayer's relation).

Specific Heat of Water

$c_{w a t e r} = 4186$ J/kg/K $= 1$ cal/g/°C. Water has the highest specific heat of common substances — this is why coastal areas have moderate climate and why water is used as a coolant.

Specific Heats of Common Substances

Substance	Specific Heat (J/kg/K)	Specific Heat (cal/g/°C)
Water	$4186$	$1.00$
Ice	$2090$	$0.50$
Steam	$2010$	$0.48$
Aluminium	$900$	$0.215$
Copper	$385$	$0.092$
Iron	$450$	$0.107$
Mercury	$140$	$0.033$

5. Heat Capacity (Thermal Capacity)

The heat capacity (or thermal capacity) of a body is the amount of heat required to raise its temperature by one degree:

$C_{b o d y} = m c$ (unit: J/K or J/°C)

Unlike specific heat (a material property), heat capacity depends on both the material AND the mass of the body. A large block of copper has more heat capacity than a small block of copper.

Water Equivalent

The water equivalent of a body is the mass of water that absorbs or releases the same amount of heat as the body for the same temperature change:

$W = \frac{m c}{c_{w a t e r}} = \frac{m c}{4200}$ kg (in SI)

Numerically, water equivalent in grams = heat capacity in cal/°C.

6. Calorimetry — Principle of Mixtures

Calorimetry is the measurement of heat changes in physical, chemical, or biological processes. The instrument used is a calorimeter — an insulated container that prevents heat exchange with the surroundings.

Principle of Calorimetry (Law of Mixtures)

When two bodies at different temperatures are mixed in a thermally insulated system:

Heat lost by hotter body = Heat gained by colder body

$m_{1} c_{1} (T_{1} - T_{m i x}) = m_{2} c_{2} (T_{m i x} - T_{2})$

where $T_{1} > T_{m i x} > T_{2}$ .

This is a direct consequence of the law of conservation of energy — assuming no heat is lost to the surroundings.

Equilibrium Temperature Formula

When two substances of masses $m_{1}$ , $m_{2}$ , specific heats $c_{1}$ , $c_{2}$ , and initial temperatures $T_{1}$ , $T_{2}$ are mixed:

$T_{m i x} = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2}}{m_{1} c_{1} + m_{2} c_{2}}$

This is a weighted average — the substance with greater thermal capacity (heat capacity = $m c$ ) dominates the final temperature.

Including the Calorimeter

In practice, the calorimeter itself absorbs heat. If the calorimeter has mass $m_{c}$ and specific heat $c_{c}$ , it is treated as another body at the initial temperature $T_{2}$ (same as the cold body placed in it):

$m_{1} c_{1} (T_{1} - T_{m i x}) = (m_{2} c_{2} + m_{c} c_{c}) (T_{m i x} - T_{2})$

7. Latent Heat

During a change of state (melting, boiling, etc.), heat is absorbed or released WITHOUT any change in temperature. This heat is called latent heat:

$Q = m L$

where $L$ is the specific latent heat (heat per unit mass for the phase change). SI unit: J/kg.

Type	Process	Symbol	For Water
Latent Heat of Fusion	Solid ↔ Liquid (melting/freezing)	$L_{f}$	$3.36 \times 10^{5}$ J/kg $= 80$ cal/g
Latent Heat of Vaporisation	Liquid ↔ Gas (boiling/condensation)	$L_{v}$	$22.6 \times 10^{5}$ J/kg $= 540$ cal/g
Latent Heat of Sublimation	Solid ↔ Gas directly	$L_{s}$	Dry ice (CO₂): $5.7 \times 10^{5}$ J/kg

Key observation: $L_{v} ≫ L_{f}$ for water ( $540$ vs $80$ cal/g). Vaporisation requires much more energy than melting because molecules must completely escape from the liquid, not just rearrange. This is why steam burns are more severe than boiling water burns at the same temperature.

Heating Curve for Water

Starting from ice at $- 10 °$ C and adding heat uniformly:

Segment 1 (ice heating): Temperature rises from $- 10 °$ C to $0 °$ C. $Q = m c_{i c e} Δ T$ . Slope is steep (small $c_{i c e}$ ).
Segment 2 (melting at 0°C): Temperature stays at $0 °$ C while $Q = m L_{f} = m \times 80$ cal/g is absorbed. Horizontal plateau.
Segment 3 (water heating): Temperature rises from $0 °$ C to $100 °$ C. $Q = m c_{w a t e r} Δ T$ . Slope is gentle (large $c_{w a t e r}$ ).
Segment 4 (boiling at 100°C): Temperature stays at $100 °$ C while $Q = m L_{v} = m \times 540$ cal/g is absorbed. Long horizontal plateau.
Segment 5 (steam heating): Temperature rises above $100 °$ C. $Q = m c_{s t e a m} Δ T$ .

The slope of each segment $= \frac{d T}{d Q} = \frac{1}{m c}$ — steeper slope means smaller specific heat.

8. Joule's Mechanical Equivalent of Heat

James Prescott Joule showed experimentally that mechanical work can be converted to heat. He defined the mechanical equivalent of heat:

$J = \frac{W}{Q} = 4.186$ J/cal

where $W$ is the mechanical work done and $Q$ is the heat produced. This established that heat and mechanical energy are both forms of energy (conservation of energy applies).

$1$ calorie $= 4.186$ joules.
$1$ kcal $= 4186$ J.
This relationship is the cornerstone connecting mechanics and thermodynamics.

Calorimetry Problems — Systematic Approach

Most calorimetry problems involve a phase change. The key is to check whether a phase change actually occurs before applying the final temperature formula.

Step	Action
1	Identify all bodies, their masses, specific heats, and initial temperatures.
2	Check if any body will undergo a phase change. Calculate heat required for the phase change ( $m L$ ). Compare with available heat from the hotter body.
3a — No phase change:	Apply: $T_{m i x} = \frac{\sum m_{i} c_{i} T_{i}}{\sum m_{i} c_{i}}$ directly.
3b — Phase change occurs:	Account for latent heat: Heat gained = $m L + m c Δ T$ . Set heat lost = heat gained and solve.
4	Verify: final temperature must be between the two initial temperatures (or at the phase change temperature if partial melting/freezing occurs).

Quick Check — Is there a Phase Change?

Heat available from hot body to cool to phase change temperature: $Q_{a v a i l} = m_{h} c_{h} (T_{h} - T_{p h a s e})$ .
Heat needed to complete phase change: $Q_{n e e d e d} = m_{c o l d} L$ .
If $Q_{a v a i l} > Q_{n e e d e d}$ → phase change completes, remaining heat raises temperature further.
If $Q_{a v a i l} < Q_{n e e d e d}$ → phase change is partial, final temperature = $T_{p h a s e}$ (e.g., mixture of ice and water at $0 °$ C).

JEE/NEET Power Tips — Avoid These Common Mistakes

Heat is NOT temperature. A large body at low temperature can have more internal energy than a small body at high temperature. Heat flows due to temperature difference, not due to internal energy difference.
$Q = m c Δ T$ gives zero for a phase change — during melting or boiling, $Δ T = 0$ but heat is still being absorbed. Always use $Q = m L$ during phase changes.
Water's specific heat is highest among common substances — approximately twice that of ice or steam. Do not assume they are equal: $c_{w a t e r} = 4186$ , $c_{i c e} = 2090$ , $c_{s t e a m} = 2010$ J/kg/K.
The final temperature cannot exceed the boiling point or go below the melting point in a simple mixing problem (at atmospheric pressure). If your answer gives $T_{m i x} > 100 °$ C for a water problem, a phase change occurred that you missed.
$L_{v}$ of water ( $540$ cal/g) is much larger than $L_{f}$ ( $80$ cal/g). Steam burns are more severe than boiling water burns because condensation releases $540$ cal/g extra. This is a direct application question in NEET.
Celsius and Fahrenheit are equal at $- 40 °$ . This is a frequently asked conceptual question in board exams — set $C = F$ in the conversion formula to derive it.
Change in temperature is the same in Celsius and Kelvin — $Δ T (° C) = Δ T (K)$ . So $c$ in J/kg/°C equals $c$ in J/kg/K numerically.

Practice Questions (JEE / NEET / Board Level)

Q1: Convert -40°F to Celsius and Kelvin.

A) -40°C, 233 K
B) -20°C, 253 K
C) -40°C, 253 K
D) 0°C, 273 K

Answer: A) -40°C, 233 K.

Explanation:

First, convert Fahrenheit to Celsius:
$C = \frac{5}{9} (F - 32) = \frac{5}{9} (- 40 - 32)$
$= \frac{5}{9} \times (- 72) = - 40 °C$

Next, convert Celsius to Kelvin:
$K = C + 273 = - 40 + 273 = 233 K$ .

Q2: 200 g of water at 80°C is mixed with 100 g of water at 20°C. The final temperature of the mixture is:

A) 50°C
B) 55°C
C) 60°C
D) 65°C

Answer: C) 60°C.

Explanation:

Since both substances are water, their specific heat capacities are the same ( $c_{1} = c_{2} = c$ ). We can use the weighted average formula for the final mixture temperature ( $T_{mix}$ ):

$T_{mix} = \frac{m_{1} T_{1} + m_{2} T_{2}}{m_{1} + m_{2}}$
$= \frac{200 \times 80 + 100 \times 20}{200 + 100}$
$= \frac{16000 + 2000}{300} = \frac{18000}{300} = 60 °C$ .

Q3: How much heat is required to convert 50 g of ice at 0°C completely to water at 100°C? ( $L_{f} = 80$ cal/g, $c_{water} = 1$ cal/g°C)

A) 4000 cal
B) 5000 cal
C) 8000 cal
D) 9000 cal

Answer: D) 9000 cal.

Explanation:

This process happens in two stages:

1. Heat required to melt the ice at 0°C:
$Q_{1} = m L_{f} = 50 \times 80 = 4000 cal$

2. Heat required to raise the temperature of the resulting water from 0°C to 100°C:
$Q_{2} = m c Δ T = 50 \times 1 \times 100 = 5000 cal$

Total heat = $Q_{1} + Q_{2} = 4000 + 5000 = 9000 cal$ .

Q4: 10 g of ice at 0°C is mixed with 10 g of water at 80°C. The final state of the mixture is:

A) All water at 0°C
B) Ice and water mixture at 0°C
C) All water at some temperature above 0°C
D) All water at 40°C

Answer: A) All water at 0°C.

Explanation:

First, calculate the maximum heat available from the hot water cooling down to 0°C:
$Q_{available} = m c Δ T = 10 \times 1 \times 80 = 800 cal$

Next, calculate the total heat required to melt all 10 g of ice:
$Q_{needed} = m L_{f} = 10 \times 80 = 800 cal$

Since the heat available exactly equals the heat needed to melt the ice ( $Q_{available} = Q_{needed}$ ), all of the ice will melt, but there will be no leftover heat to raise the temperature of the water. The final state is 20 g of water at exactly 0°C.

Q5: The temperature of a body on the Kelvin scale is 500 K. What is this temperature on the Celsius and Fahrenheit scales?

A) 227°C and 440.6°F
B) 227°C and 220°F
C) 500°C and 932°F
D) 273°C and 523°F

Answer: A) 227°C and 440.6°F.

Explanation:

First, convert Kelvin to Celsius:
$C = K - 273 = 500 - 273 = 227 °C$

Next, convert Celsius to Fahrenheit:
$F = (\frac{9}{5} \times C) + 32$
$F = (\frac{9 \times 227}{5}) + 32$
$F = \frac{2043}{5} + 32 = 408.6 + 32 = 440.6 °F$ .

Q6: A piece of iron of mass 100 g and specific heat 0.45 J/g°C is placed in 200 g of water at 20°C. If the iron is at 100°C initially, find the final equilibrium temperature. (Specific heat of water = 4.2 J/g°C)

A) 28.4°C
B) 24.1°C
C) 30.2°C
D) 35.8°C

Answer: B) 24.1°C.

Explanation:

By the principle of calorimetry, the heat lost by the iron equals the heat gained by the water.

$m_{Fe} c_{Fe} (T_{Fe} - T_{f}) = m_{w} c_{w} (T_{f} - T_{w})$

Substitute the given values:
$100 \times 0.45 \times (100 - T_{f})$
$= 200 \times 4.2 \times (T_{f} - 20)$
$45 (100 - T_{f}) = 840 (T_{f} - 20)$
$4500 - 45 T_{f} = 840 T_{f} - 16800$

Group the terms to solve for $T_{f}$ :
$4500 + 16800 = 840 T_{f} + 45 T_{f}$
$21300 = 885 T_{f}$
$T_{f} = \frac{21300}{885} \approx 24.067 °C$

The equilibrium temperature is approximately 24.1°C.

Q7 (JEE Advanced type): 100 g of ice at -10°C is heated. The total heat required to convert it completely to steam at 100°C is: ( $c_{ice} = 0.5$ cal/g°C, $L_{f} = 80$ cal/g, $c_{water} = 1$ cal/g°C, $L_{v} = 540$ cal/g)

A) 72050 cal
B) 62050 cal
C) 72500 cal
D) 71500 cal

Answer: C) 72500 cal.

Explanation:

This process involves four distinct stages of heating and phase changes:

1. Heat ice from -10°C to 0°C:
$Q_{1} = m c_{ice} Δ T = 100 \times 0.5 \times 10 = 500 cal$

2. Melt ice at 0°C to water at 0°C:
$Q_{2} = m L_{f} = 100 \times 80 = 8000 cal$

3. Heat water from 0°C to 100°C:
$Q_{3} = m c_{water} Δ T = 100 \times 1 \times 100$
$= 10000 cal$

4. Vaporize water at 100°C to steam at 100°C:
$Q_{4} = m L_{v} = 100 \times 540 = 54000 cal$

Total Heat required:
$Q_{total} = 500 + 8000 + 10000 + 54000$
$= 72500 cal$ .

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Temperature, Heat & Calorimetry

1. Temperature and Its Measurement

Zeroth Law of Thermodynamics

Temperature Scales

Temperature Conversion Formulae

Key Conversion Results

2. Ideal Gas Temperature Scale and Absolute Zero

3. Heat and Thermal Energy

Distinction: Heat vs Temperature vs Internal Energy

4. Specific Heat Capacity

Molar Heat Capacity

Specific Heat at Constant Pressure and Constant Volume

Specific Heat of Water

Specific Heats of Common Substances

5. Heat Capacity (Thermal Capacity)

Water Equivalent

6. Calorimetry — Principle of Mixtures

Principle of Calorimetry (Law of Mixtures)

Equilibrium Temperature Formula

Including the Calorimeter

7. Latent Heat

Heating Curve for Water

8. Joule's Mechanical Equivalent of Heat

Calorimetry Problems — Systematic Approach

Quick Check — Is there a Phase Change?

Practice Questions (JEE / NEET / Board Level)

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Practice Questions

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