### Solution:
Given:
1. Mass of water (m): 335 grams
2. Initial temperature (T_initial): 65.5°C
3. Heat lost (Q): 9750 Joules
4. Specific heat of water (c): 4.186 J/g°C
Objective: Calculate the final temperature of the water after losing the heat.
### Step-by-Step Solution:
1. Understand the relationship:
The formula relating heat transfer (Q), mass of the substance (m), specific heat (c), and change in temperature (ΔT) is:
[tex]\( Q = m \cdot c \cdot \Delta T \)[/tex]
Where:
- [tex]\( Q \)[/tex] is the heat energy (in Joules)
- [tex]\( m \)[/tex] is the mass (in grams)
- [tex]\( c \)[/tex] is the specific heat capacity (in J/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C)
2. Rearrange the formula to solve for the change in temperature (ΔT):
[tex]\( \Delta T = \frac{Q}{m \cdot c} \)[/tex]
3. Substitute the known values into the formula:
[tex]\( \Delta T = \frac{9750 \text{ Joules}}{335 \text{ g} \cdot 4.186 \text{ J/g°C}} \)[/tex]
4. Calculate the change in temperature (ΔT):
[tex]\[ \Delta T = \frac{9750}{335 \times 4.186} \][/tex]
[tex]\[ \Delta T = \frac{9750}{1402.31} \][/tex]
[tex]\[ \Delta T \approx 6.9528°C \][/tex]
5. Determine the final temperature:
Since the water is losing heat, the final temperature will be lower than the initial temperature. Thus,
[tex]\( T_{\text{final}} = T_{\text{initial}} - \Delta T \)[/tex]
Substitute the values:
[tex]\( T_{\text{final}} = 65.5°C - 6.9528°C \)[/tex]
[tex]\( T_{\text{final}} \approx 58.5472°C \)[/tex]
### Conclusion:
The change in temperature (ΔT) is approximately [tex]\( 6.9528°C \)[/tex], and the final temperature of the water after losing 9750 J of heat is approximately [tex]\( 58.5472°C \)[/tex].
