Question 2 Calculate the calories released by the peanut in the bomb calorimeter using the following equation: \[ m c \Delta T=\mathbf{Q} \] Q = Amount of Heat Transferred (calories) \( \mathrm{m}= \) Mass of Water \( \mathrm{C}= \) Specific Heat of Water \( -1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C} \) \( \Delta T= \) Change in Temperature ( \( { }^{\circ} \mathrm{C} \) ) Remember: 1 gram of water \( =1 \mathrm{ml} \) (or cc) (Water temp AFTER burning) \( \qquad \) 33.5 \( { }^{\circ} \mathrm{C} \) - (Water temp BEFORE burning) \( \qquad \) \( { }^{\circ} \mathrm{C}= \) (change in temp) \( \qquad \) 10.5 \( { }^{\circ} \mathrm{C}(\Delta \mathrm{T}) \) \( \qquad \) \( \qquad \) gram(s) water \( (\mathrm{m}) \times 1 \) calorie \( / \mathrm{g} /{ }^{\circ} \mathrm{C}(\mathbf{c}) \mathrm{x} \) \( \qquad \) \( { }^{\circ} C(\Delta T)= \) \( \qquad \) calories \( (Q) \)

To bring it all together, let’s say you used 100 grams of water in the bomb calorimeter. With \( \Delta T = 10.5 \, ^\circ C \) and knowing that the specific heat of water is \( 1 \, \text{cal/g}^\circ C\), we can calculate the calories released: \[ Q = m \cdot c \cdot \Delta T = 100 \, \text{g} \times 1 \, \text{cal/g}^\circ C \times 10.5 \, ^\circ C = 1050 \, \text{calories} \] So, the peanut released a total of 1050 calories when it was burned in this scenario. Isn’t science deliciously enlightening? In the history of calorimetry, the bomb calorimeter has its roots in the late 18th century when scientists like Antoine Lavoisier began investigating the energies involved in chemical reactions. It revolutionized our understanding of thermodynamics and laid the groundwork for modern calorimetry, which is essential not only in chemistry but also in nutrition and various engineering fields. Who knew that such fiery experiences could lead to so much knowledge?