A hot iron ball is dropped into a 200. g sample of water initially at 50.° C. If 8.4 kJ of heat is t — Thermodynamics Chemistry Question

STEPS:

1. Identify the relevant thermodynamic equation: The heat absorbed or released by a substance during a temperature change without a phase transition is described by the calorimetry equation:
$$q = mc\Delta T$$
where $q$ is the heat transferred, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature ($T_f - T_i$).
2. Identify the given variables for the water:
* Mass of the water ($m$) = $200.\text{ g}$
* Initial temperature ($T_i$) = $50.^\circ\text{C}$
* Specific heat capacity of water ($c$) = $4.2\text{ J/(g}\cdot^\circ\text{C)}$
* Heat transferred to the water ($q$) = $8.4\text{ kJ}$
3. Convert the heat energy to matching units: The specific heat capacity is given in Joules ($\text{J}$), while the transferred heat is given in kilojoules ($\text{kJ}$). To perform the calculation, convert the heat from $\text{kJ}$ to $\text{J}$:
$$q = 8.4\text{ kJ} \times \frac{1000\text{ J}}{1\text{ kJ}} = \mathbf{8400\text{ J}}$$
4. Calculate the temperature change ($\Delta T$) of the water: Rearrange the calorimetry equation to solve for $\Delta T$:
$$\Delta T = \frac{q}{mc}$$
Substitute the values into the equation:
$$\Delta T = \frac{8400\text{ J}}{(200.\text{ g}) \times (4.2\text{ J/(g}\cdot^\circ\text{C)})} = \frac{8400}{840} = \mathbf{10.^\circ\text{C}}$$
5. Determine the final temperature ($T_f$): Because heat is transferred *from* the hot iron ball *to* the water, the water is absorbing heat. This means its temperature must increase:
$$\Delta T = T_f - T_i \implies T_f = T_i + \Delta T$$
$$T_f = 50.^\circ\text{C} + 10.^\circ\text{C} = \mathbf{60.^\circ\text{C}}$$
This matches Option C.

*

WHY_OTHERS_WRONG:

* Option A is incorrect (40.°C): This value is obtained if a student correctly calculates the temperature change as $10.^\circ\text{C}$ but mistakenly subtracts it from the initial temperature ($50 - 10 = 40$). Since heat is transferred *to* the water, its temperature must rise, not fall.
* Option B is incorrect (51°C): This mistake occurs if a student fails to convert the heat value from kilojoules to Joules. Using $8.4\text{ J}$ instead of $8400\text{ J}$ yields an incorrect temperature change:
$$\Delta T = \frac{8.4\text{ J}}{840\text{ J/}^\circ\text{C}} = 0.01^\circ\text{C}$$
Adding this to $50.^\circ\text{C}$ and rounding inappropriately can lead to selecting $51^\circ\text{C}$.
* Option D is incorrect (70.°C): This is a distractor value that could result from a calculation error (such as doubling the temperature change to $20.^\circ\text{C}$) or applying an incorrect specific heat value.