A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperat — Thermodynamics Chemistry Question

STEPS:

1. Recall the fundamental equation for calorimetry: The heat energy ($q$) absorbed by the water in a calorimeter is calculated using the formula:
$$q = m c \Delta T$$
where $m$ is the mass of the water, $c$ is the specific heat capacity of water ($4.2\text{ J/(g}\cdot^\circ\text{C)}$), and $\Delta T$ is the change in the water's temperature.
2. Calculate the heat released by the combustion of the cashew sample:
* Mass of water ($m$) = $1000.\text{ g}$
* Temperature change ($\Delta T$) = $25.0^\circ\text{C} - 20.0^\circ\text{C} = 5.0^\circ\text{C}$
* Heat absorbed by the water ($q_{\text{cashew}}$):
$$q = (1000.\text{ g}) \times \left(4.2\text{ J/(g}\cdot^\circ\text{C)}\right) \times (5.0^\circ\text{C}) = 21,000\text{ J} = \mathbf{21\text{ kJ}}$$
* Since this energy was produced by burning exactly $1.0\text{ g}$ of cashew, the energy content per gram of cashew is:
$$\text{Energy content of cashew} = \frac{21\text{ kJ}}{1.0\text{ g}} = \mathbf{21\text{ kJ/g}}$$
3. Calculate the heat released by the combustion of the marshmallow sample:
* Mass of water ($m$) = $2000.\text{ g}$
* Temperature change ($\Delta T$) = $30.0^\circ\text{C} - 25.0^\circ\text{C} = 5.0^\circ\text{C}$
* Heat absorbed by the water ($q_{\text{marshmallow}}$):
$$q = (2000.\text{ g}) \times \left(4.2\text{ J/(g}\cdot^\circ\text{C)}\right) \times (5.0^\circ\text{C}) = 42,000\text{ J} = \mathbf{42\text{ kJ}}$$
* Since this energy was produced by burning $3.0\text{ g}$ of marshmallow, calculate the energy content per gram:
$$\text{Energy content of marshmallow} = \frac{42\text{ kJ}}{3.0\text{ g}} = \mathbf{14\text{ kJ/g}}$$
4. Compare the energy released per gram of each substance:
* $1.0\text{ g}$ of cashew releases $21\text{ kJ}$ of energy.
* $1.0\text{ g}$ of marshmallow releases $14\text{ kJ}$ of energy.
* Comparing these values, the combustion of $1.0\text{ g}$ of cashew releases more energy than the combustion of $1.0\text{ g}$ of marshmallow ($21\text{ kJ/g} > 14\text{ kJ/g}$), which confirms Option C is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option states that the cashew releases less energy per gram than the marshmallow. This is a common trap if a student only looks at the total heat released by the entire sample of marshmallow ($42\text{ kJ}$) versus the cashew ($21\text{ kJ}$), while failing to divide the marshmallow's heat by its mass of $3.0\text{ g}$ to get the energy *per gram*.
• Option B is incorrect: This option claims that both substances release the same amount of energy per gram. This would only be true if the marshmallow's combustion released $63\text{ kJ}$ of total heat (yielding $21\text{ kJ/g}$).
• Option D is incorrect: While the two systems started with different parameters (water masses, food masses, and initial temperatures), a direct and valid scientific comparison can absolutely be made. Using the calorimetry equation allows us to normalize the data to find the specific energy density (energy per unit mass, $\text{kJ/g}$) for each food item.