Understanding Enthalpy and Calorimetric Measurements in Combustion and Heat Transfer

Introduction

In the realm of thermodynamics, understanding enthalpy and calorimetric measurements is crucial for solving practical problems involving heat transfer and reactions. This article delves into two illustrative examples: the combustion of methane to heat water and the transfer of heat between a heated silver coin and water. These problems not only solidify core concepts but also enhance problem-solving skills related to heat energy calculations in chemical reactions and thermal systems.

Combustion of Methane

Problem Statement

The first problem challenges us to determine the mass of methane (CH4) that must be combusted to increase the temperature of 1 kg of water from 25°C to 100°C.

Key Information

The combustion reaction:
CH4 + 2 O2 → CO2 + 2 H2O
Standard enthalpy change (ΔH) for this reaction: -890 kJ/mol
Specific heat capacity of water: 4.18 J/g°C

Step-by-Step Solution

Calculate Heat Required to Heat the Water:
We will use the formula for heat transfer:
[ Q = mc\Delta T ]
where
- ( m ) = mass of water (1000 g, since 1 kg = 1000 g)
- ( c ) = specific heat capacity of water (4.18 J/g°C)
- ( \Delta T = T_{final} - T_{initial} = 100°C - 25°C = 75°C )
  Substituting in these values:
  [ Q_{water} = (1000 g)(4.18 J/g°C)(75°C) = 313500 J = 313.5 kJ ]
This indicates that to heat the water, 313.5 kJ of heat energy is required.
Relate Heat Released to Heat Required:
In an exothermic reaction like methane combustion, the heat released (Q_reaction) is equal and opposite to the heat absorbed by the water:
[ Q_{reaction} = -Q_{water} = -313.5 kJ ]
Calculate Moles of Methane Required:
From the enthalpy change, we know that: [ 890 kJ ] is released for every mole of methane combusted.
To find the moles of methane required, we perform the calculation:
[ n = \frac{-Q_{reaction}}{ΔH} = \frac{-313.5 kJ}{-890 kJ/mol} = 0.352 moles ]
Determine Mass of Methane:
The molar mass of methane (CH4) is approximately 16.04 g/mol. Thus, the mass of methane is calculated as follows:
[ m_{methane} = n \times molar mass = 0.352 moles \times 16.04 g/mol = 5.63 g ]

Therefore, approximately 5.63 g of methane must be combusted to heat 1 kg of water from 25°C to 100°C.

Heat Transfer Between Silver Coin and Water

Problem Statement

In the second scenario, we need to determine the final temperature of a system where a 25 g silver coin heated to 45°C is placed in 25 g of water at 22°C.

Key Information

Mass of silver coin: 25 g
Initial temperature of silver (T1): 45°C
Initial temperature of water (T2): 22°C
Specific heat capacity of silver: 0.24 J/g°C
Specific heat capacity of water: 4.18 J/g°C

Analysis

Understanding Heat Transfer:
The heat lost by the silver coin equals the heat gained by the water. [ Q_{silver} = -Q_{water} ]
Setting Up the Equations:
Using the specific heat capacity formula for each substance, we have:
- For silver:
  [ Q_{silver} = m_{silver} \cdot c_{silver} \cdot (T_{final} - T1) ]
- For water:
  [ Q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T2) ]
Since the masses are equal (25 g), we can simplify: [ 25g imes 0.24 J/g°C imes (T_f - 45°C) = - (25g imes 4.18 J/g°C imes (T_f - 22°C)) ]
Solving for Final Temperature:
By balancing the heat lost and gained, we rearrange the equation to isolate ( T_f ). After calculation (details omitted for brevity), we find: [ T_f = 23.2°C ]

Conclusion

The final temperature, 23.2°C, is indeed closer to the initial temperature of water (22°C) than that of the silver coin (45°C) due to the significant difference in their specific heat capacities.

Summary

In this article, we explored the concepts of enthalpy and calorimetric measurements through detailed calculations involving methane combustion and heat transfer between a heated silver coin and water. We learned that understanding heat transfer and enthalpy changes is essential for both theoretical problems and practical applications in thermodynamics. With a firm grasp of these concepts, we can approach similar problems with confidence and clarity.