Calculate the amount of energy (in J) gained by the cool water when it was mixed with the warm water at 4 minutes. ### Calorimeter Calibration#### Graph AnalysisThe graph displays the temperature change over time for three different trials labeled A, B, and C. The x-axis represents time in minutes, ranging from 0 to 8 minutes. The y-axis shows temperature in degrees Celsius from 0 to 50°C.- **Trial A (Red Squares)**: The equation of the trend line is \( y = -0.44x + 45.86 \). This line shows a gradual decrease in temperature, indicating that 50 mL of warm water is slowly cooling over time.- **Trial B (Green Triangles)**: The equation for this trend line is \( y = -0.35x + 34.533 \). This represents the temperature of a mixture of warm and cool water, which also shows a gradual cooling over time.- **Trial C (Blue Diamonds)**: The trend line equation is \( y = -0.04x + 23.41 \), depicting a relatively stable temperature. This indicates that 50 mL of cool water stays relatively constant over the observed period.#### Additional Notes- **Red**: Temperature as 50 mL of warm water slowly cooled over time.- **Green**: Temperature as a mixture of both warm and cool water slowly cooled over time.- **Blue**: Temperature as 50 mL of cool water stayed relatively constant over time.#### Formula for Calculating EnergyThe amount of energy (in Joules) can be calculated using the formula:\[ \text{Energy (J)} = \text{mass} \times C_{\text{water}} \times \text{change in temp} \]This section provides a comprehensive understanding of how the temperature of different water samples changes over time in a calorimeter setup, useful for educational purposes in thermodynamics or chemistry experiments.

Given: Volume of water = 50 mL Time = 4 min From the calibration curve at 4 min, the temperature of cool water = 23.25 oC From the calibration curve at 4 min, the final temperature of water = 33.133 oC Let the density of water be 1 g/mLCalculation for mass of water: mass=volume×densitymass=50 mL×1 g/mLmass = 50 gCalculation for amount of energy required to heat water: Q=mC∆TQ=50 g×4.184 J/g.°C×(33.133 °C-23.25 °C)Q=50 g×4.184 J/g.°C×9.883 °CQ=2067.5 J=2.07 kJ