When 5.0 g of NH4ClO4(s) is added to 100. mL of water in a calorimeter, the temperature of the solut — Thermodynamics Chemistry Question

STEPS:

1. Identify the constant variable between both trials:
In both trials, exactly $5.0\text{ g}$ of the solid salt $\text{NH}_4\text{ClO}_4$ is dissolved in water. Since the chemical process and the mass of the reactant are identical, the total amount of heat absorbed during the dissolution ($q$) must be identical in both calorimeters.
2. Recall the calorimetry heat equation:
The heat exchanged by the solution is given by:
$$q = m \cdot c \cdot \Delta T$$
where $m$ is the mass of the solution, $c$ is the specific heat capacity of the solution (which is approximately equal to that of pure water, $4.18\text{ J/g}^\circ\text{C}$), and $\Delta T$ is the temperature change. Because the density of water is $1.0\text{ g/mL}$, the mass of the solvent dominates the solution mass:
* First calorimeter: $100.\text{ mL} \approx 100.\text{ g}$ of water
* Second calorimeter: $1000.\text{ mL} \approx 1000.\text{ g}$ of water
3. Establish the relationship between mass and temperature change:
Since the heat absorbed ($q$) and specific heat capacity ($c$) are constant, the temperature change is inversely proportional to the mass of the water:
$$m_1 \cdot \Delta T_1 = m_2 \cdot \Delta T_2$$
4. Calculate the new temperature change ($\Delta T_2$):
The volume (and therefore mass) of the water has increased by a factor of 10 (from $100.\text{ mL}$ to $1000.\text{ mL}$). Because there is 10 times as much water to absorb the same quantity of heat, the magnitude of the temperature change must be 10 times smaller:
$$\Delta T_2 = \frac{\Delta T_1}{10} = \frac{-3.0^\circ\text{C}}{10} = -0.3^\circ\text{C}$$
5. Determine the final temperature:
Since the dissolution of ammonium perchlorate is endothermic (indicated by a temperature *decrease*), the temperature of the water will drop by $0.3^\circ\text{C}$ from its starting temperature of $25.0^\circ\text{C}$:
$$T_{\text{final}} = T_{\text{initial}} + \Delta T_2 = 25.0^\circ\text{C} - 0.3^\circ\text{C} = \mathbf{24.7^\circ\text{C}}$$
This aligns perfectly with Option B.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (22.0°C): This value is obtained by subtracting the original $3.0^\circ\text{C}$ temperature drop directly from the new initial temperature ($25.0^\circ\text{C} - 3.0^\circ\text{C}$). A student would choose this if they incorrectly assumed that dissolving the same amount of solute always results in the identical temperature change, ignoring the fact that a larger volume of water has a much higher overall heat capacity and resists temperature changes more.
• Option C is incorrect (25.3°C): This represents a temperature *increase* of $0.3^\circ\text{C}$ ($25.0^\circ\text{C} + 0.3^\circ\text{C}$). Although the student scaled the magnitude of the temperature change correctly for the $10\times$ larger volume, they mistakenly treated the dissolution process as exothermic rather than endothermic.
• Option D is incorrect (28.0°C): This represents a temperature *increase* of $3.0^\circ\text{C}$ ($25.0^\circ\text{C} + 3.0^\circ\text{C}$). Choosing this option requires making two errors: failing to scale the temperature change to account for the larger volume of water, and misinterpreting the endothermic process as exothermic.