Calorimetry

Mathematical approach:

1. Identify heat lost by water cooling from 25°C to 0°C: $Q_{water} = m_{water} \times c_{water} \times (25°C - 0°C)$ $Q_{water} = m_{water} \times c_{water} \times 25°C$

2. This heat is used to melt a portion of ice (mass $m_{melted}$): $Q_{ice} = m_{melted} \times L_f$ where $L_f$ is the latent heat of fusion of ice

3. By calorimetry principle: $Q_{water} = Q_{ice}$ $m_{water} \times c_{water} \times 25°C = m_{melted} \times L_f$

4. Solve for $m_{melted}$: $m_{melted} = \frac{m_{water} \times c_{water} \times 25°C}{L_f}$

Example: If 200g water cools from 25°C to 0°C, it releases 21,000J, which can melt approximately 62g of ice.

The principle of calorimetry states that the total heat given by hot objects equals the total heat received by cold objects (assuming no heat exchange with surroundings).

Mathematically: $Q_{lost} = Q_{gained}$

For multiple objects:

• $\sum_{hot} m_i c_i (T_{initial,i} - T_{final}) = \sum_{cold} m_j c_j (T_{final} - T_{initial,j})$

Example: When a hot metal piece at 90°C is placed in cold water at 20°C in an insulated calorimeter, heat flows until equilibrium is reached. The heat lost by the metal equals the heat gained by the water and calorimeter.

Note: This principle assumes perfect thermal isolation from the environment.

Critical measurements:

1. Mass of the solid ($m_1$)
2. Mass of calorimeter and stirrer ($m_2$)
3. Mass of water ($m_3$)
4. Initial temperature of solid ($θ_1$)
5. Initial temperature of calorimeter, stirrer and water ($θ_2$)
6. Final equilibrium temperature of mixture ($θ$)

Calculation formula: $s_1 = \frac{(m_2s_2 + m_3s_3)(θ - θ_2)}{m_1(θ_1 - θ)}$

Where:

• $s_1$ = specific heat capacity of solid (unknown)
• $s_2$ = specific heat capacity of calorimeter material
• $s_3$ = specific heat capacity of water

Process requires:

• Heating solid in steam chamber until temperature stabilizes
• Transferring quickly to calorimeter
• Measuring temperature change accurately
• Minimizing heat loss to environment

Searle's Cone Method determines J by converting mechanical work to heat:

Setup:

• Inner conical vessel filled with water
• Outer vessel rotates, causing friction
• Friction torque balanced by suspended weights
• Temperature rise measured

Calculation: $J = \frac{2\pi n Mgr}{(m_1s_1 + m_2s_2)(θ_2 - θ_1)}$

Where:

• $n$ = number of rotations
• $M$ = mass on hanging pan
• $g$ = acceleration due to gravity
• $r$ = radius of disc
• $m_1$ = mass of water
• $m_2$ = mass of vessels
• $s_1, s_2$ = specific heat capacities
• $θ_1, θ_2$ = initial and final temperatures

Physical principle: This demonstrates energy conservation - mechanical work (rotation against friction) converts completely to thermal energy (temperature increase), establishing the equivalence between work and heat.

Experimental value: $J = 4.186$ joules/calorie

Analysis process:

1. Determine possible heat transfers in sequence:

   • Heat to warm ice to 0°C: $Q_1 = m_{ice} \times c_{ice} \times (0°C - T_{ice})$
   • Heat to melt ice at 0°C: $Q_2 = m_{ice} \times L_f$
   • Heat to warm melted water: $Q_3 = m_{ice} \times c_{water} \times (T_{final} - 0°C)$
   • Heat lost by warm water: $Q_4 = m_{water} \times c_{water} \times (T_{water} - T_{final})$

2. Compare available heat with required heat:

   • If $Q_4 < Q_1$: Final temperature below 0°C, no melting
   • If $Q_1 < Q_4 < (Q_1 + Q_2)$: Final temperature = 0°C, partial melting
   • If $Q_4 > (Q_1 + Q_2)$: Complete melting, final temperature above 0°C

3. For case of partial melting, calculate mass melted: $m_{melted} = \frac{Q_4 - Q_1}{L_f}$

Example: For 5g ice at -20°C added to 10g water at 30°C, first calculate heat available from water cooling to 0°C (1260J) and compare to heat needed to warm ice to 0°C (210J) plus heat to melt all ice (1680J).

Experimental measurement:

1. Setup components:

   • Steam generator/boiler
   • Steam trap to ensure dry steam
   • Calorimeter with water at known temperature
   • Accurate thermometers and weighing balance

2. Procedure:

   • Pass steam into calorimeter containing water
   • Measure initial and final temperatures
   • Determine mass of condensed steam

3. Calculation formula: $L = \frac{(m_1s_1 + m_2s_2)(θ_3 - θ_2)}{m_3} - s_2(θ_1 - θ_3)$

   Where:

   • $m_1$ = mass of calorimeter
   • $m_2$ = mass of water in calorimeter
   • $m_3$ = mass of condensed steam
   • $θ_1$ = temperature of steam
   • $θ_2, θ_3$ = initial and final temperatures of calorimeter

4. Necessary corrections:

   • Subtract heat released by cooling condensed water from steam temperature to final temperature
   • Account for heat lost to surroundings
   • Correct for heat capacity of calorimeter and stirrer
   • Ensure steam is dry (no water droplets)

Example: If 1.5g of steam at 100°C condenses in a calorimeter, raising water temperature from 25°C to 30°C, both the heat of condensation and the heat released during cooling from 100°C to 30°C must be accounted for.

Temperature plateaus occur because all added heat goes into phase change rather than temperature increase:

Characteristics:

• During phase transitions (melting, boiling), temperature remains constant
• Plateau duration proportional to mass and inversely proportional to heating rate
• Slope changes occur at phase transition temperatures

Information extractable:

• Melting/boiling points (plateau temperature)
• Latent heat (from plateau duration): $L = \frac{Q \times t_{plateau}}{m}$ where Q is heat input rate
• Relative quantities of substances (from plateau lengths)
• Purity of substances (pure substances have sharp, flat plateaus)

Example: A temperature-time graph for ice heated at constant rate shows:

1. Linear increase from -20°C to 0°C (solid phase)
2. Plateau at 0°C (melting)
3. Linear increase from 0°C to 100°C (liquid phase)
4. Plateau at 100°C (vaporization)
5. Linear increase above 100°C (gas phase)

To calculate the specific heat capacity of a solid using Regnault's apparatus data:

1. Rearrange the heat exchange equation to solve for s₁: s₁ = [(m₂s₂ + m₃s₃)(θ - θ₂)]/[m₁(θ₁ - θ)]

Where:

• s₁: specific heat capacity of the solid (unknown)
• m₁: mass of the solid
• θ₁: initial temperature of the solid (typically 100°C)
• m₂, s₂: mass and specific heat capacity of the calorimeter
• m₃, s₃: mass and specific heat capacity of water (4186 J/kg·K)
• θ₂: initial temperature of calorimeter and water
• θ: final equilibrium temperature

2. Substitute all known values into the equation and solve for s₁

Example: If a 75g aluminum sample cooled from 100°C to 28°C while 150g of water warmed from 20°C to 28°C in a 50g copper calorimeter (s = 389 J/kg·K), the calculated specific heat capacity of aluminum would be approximately 900 J/kg·K.

Note: For accurate results, minimize heat loss to surroundings by using a well-insulated calorimeter and conducting the transfer quickly.

The principle involves transferring heat from condensing steam to cold water and measuring the resulting temperature change:

1. Steam at 100°C condenses to water at 100°C, releasing latent heat (L)
2. The condensed water cools from 100°C to the final mixture temperature, releasing sensible heat
3. The calorimeter and cold water absorb this heat, increasing in temperature
4. Using conservation of energy: heat lost by steam = heat gained by cold system

This is mathematically expressed as: m₃L + m₃s₂(θ₁ - θ₃) = m₁s₁(θ₃ - θ₂) + m₂s₂(θ₃ - θ₂)

Where:

• m₃ = mass of condensed steam
• L = specific latent heat of vaporization
• θ₁ = steam temperature (100°C)
• θ₃ = final temperature
• θ₂ = initial cold water temperature
• s₁, s₂ = specific heat capacities of calorimeter material and water

Solving for L gives the specific latent heat of vaporization.

Searle's Cone Method is an experimental technique used to measure the mechanical equivalent of heat (J), which represents how much mechanical work is equivalent to a unit of heat energy.

Key components and principles:

• Uses two conical vessels (inner and outer) with friction between them
• Outer vessel rotates while inner vessel remains stationary due to balanced torques
• Work done by friction converts to heat, raising temperature of water in inner vessel
• The method calculates J using the formula: $J = \frac{2\pi n Mgr}{(m_1s_1 + m_2s_2)(θ_2 - θ_1)}$ where n is number of rotations, M is mass on pan, r is radius, m₁ and m₂ are masses of water and vessels, s₁ and s₂ are specific heat capacities, and θ₁ and θ₂ are initial and final temperatures.

The accepted value is 4.186 joules per calorie, confirming that heat is a form of energy.