Sure! Let's solve this step-by-step.
### Step 1: Given values
- Mass of water (m): 1299.1 grams
- Heat absorbed (Q): 65.12 kJ
- Initial temperature (T_initial): 7°C
- Specific heat capacity of water (c): [tex]\(4.18 \, \text{J/g°C}\)[/tex]
### Step 2: Convert heat absorbed from kJ to J
We know that [tex]\( 1 \, \text{kJ} = 1000 \, \text{J} \)[/tex]. So we convert the heat absorbed from kJ to J:
[tex]\[ Q = 65.12 \, \text{kJ} \times 1000 \, \frac{\text{J}}{\text{kJ}} = 65120 \, \text{J} \][/tex]
### Step 3: Calculate the change in temperature
We use the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
Where [tex]\( \Delta T \)[/tex] is the change in temperature.
Solving for [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{Q}{m \cdot c} \][/tex]
Substitute the given values:
[tex]\[ \Delta T = \frac{65120 \, \text{J}}{1299.1 \, \text{g} \times 4.18 \, \text{J/g°C}} \][/tex]
[tex]\[ \Delta T \approx 11.99°C \][/tex]
### Step 4: Calculate the final temperature
We add the change in temperature to the initial temperature to find the final temperature:
[tex]\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \][/tex]
[tex]\[ T_{\text{final}} = 7°C + 11.99°C \][/tex]
[tex]\[ T_{\text{final}} \approx 18.99°C \][/tex]
### Summary
Thus, after absorbing 65.12 kJ of heat, the final temperature of the water sample will be approximately 18.99°C.
