The problem describes a situation where 8.00 g of NaOH is dissolved in 280.0 g of water, and after the dissolution, the temperature of the water rises to 32.0°C. The heat absorbed by the water is 6720 J, and it takes 4.00 J to raise the temperature of 1.00 g of water by 1.00°C. The goal is to find the initial temperature of the water.

Let's use the specific heat formula:

$q = mc\Delta T$

Where:

• $q$ is the heat absorbed (6720 J),
• $m$ is the mass of water (280.0 g),
• $c$ is the specific heat capacity of water (4.00 J/g°C),
• $\Delta T$ is the change in temperature.

We need to find the initial temperature $T_{\text{initial}}$, knowing that the final temperature $T_{\text{final}}$ is 32.0°C.

First, solve for $\Delta T$:

$\Delta T = \frac{q}{mc} = \frac{6720 \, \text{J}}{280.0 \, \text{g} \times 4.00 \, \text{J/g°C}} = \frac{6720}{1120} \, \text{°C} = 6.0 \, \text{°C}$

Now, use the change in temperature to find the initial temperature:

$\Delta T = T_{\text{final}} - T_{\text{initial}}$

$6.0 \, \text{°C} = 32.0 \, \text{°C} - T_{\text{initial}}$

$T_{\text{initial}} = 32.0 \, \text{°C} - 6.0 \, \text{°C} = 26.0 \, \text{°C}$

The initial temperature of the water is 26.0°C.

Would you like more details on this calculation or have any other questions?

Here are 5 related questions:

1. How does the specific heat capacity of a substance affect the heat absorbed or released during a temperature change?
2. If the mass of water was doubled, how would it affect the final temperature?
3. How would the result change if the specific heat capacity of water were different?
4. What if the NaOH was dissolved in a different liquid with a different specific heat capacity—how would that impact the outcome?
5. How can calorimetry be used to determine the enthalpy change of a chemical reaction?

Tip: In calorimetry, keeping track of units and the specific heat capacity is crucial for accurate calculations.