Chapter 5: Problem 38
A 0.692 -g sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) was burned in a constant volume calorimeter. The temperature rose from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C} .\) The calorimeter contained 575 g of water, and the bomb had a heat capacity of \(650 \mathrm{J} / \mathrm{K} .\) What is \(\Delta U\) per mole of glucose?
Short Answer
Step by step solution
Calculate Temperature Change
Calculate Heat Absorbed by Water
Calculate Heat Absorbed by Bomb Calorimeter
Total Heat Released by Combustion Reaction
Convert Heat to Molar Basis
Provide Final Expression for \(\Delta U\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
Specific Heat Capacity
In the exercise, water in the calorimeter absorbs heat from the reaction, and its specific heat capacity plays a critical role in determining the total heat change. This allows us to quantify the energy released by the glucose combustion.
Molar Mass
When solving calorimetry problems, accurately knowing the molar mass ensures the correct interpretation of energy changes on a molar level, which is often the standard unit of measurement in energy calculations.
Temperature Change
This increase in temperature is recorded and used to calculate the total heat absorbed by both the water and the calorimeter. The overall heat change is split into contributions from these two sources to precisely determine how much energy was released per mole of glucose. This process highlights the importance of precise temperature measurements in energy calculations.