Sure, let's break down the steps to calculate how much energy is lost by the hot water when 1000 grams of water at 100°C is mixed with 2000 grams of water at 2°C, resulting in a final temperature of 34.67°C.
### Given Data:
- Mass of hot water ([tex]\( m_{\text{hot}} \)[/tex]) = 1000 grams
- Initial temperature of hot water ([tex]\( T_{\text{hot initial}} \)[/tex]) = 100°C
- Final temperature of the mixture ([tex]\( T_{\text{final}} \)[/tex]) = 34.67°C
- Specific heat capacity of water ([tex]\( c \)[/tex]) = 4.18 J/g°C
### Energy Lost by Hot Water
To find the energy lost by the hot water, we can use the formula for heat transfer:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat energy
- [tex]\( m \)[/tex] is the mass
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature
First, calculate the change in temperature ([tex]\( \Delta T \)[/tex]) for the hot water:
[tex]\[ \Delta T = T_{\text{hot initial}} - T_{\text{final}} \][/tex]
[tex]\[ \Delta T = 100°C - 34.67°C \][/tex]
[tex]\[ \Delta T = 65.33°C \][/tex]
Now, plug the values into the formula:
[tex]\[ Q_{\text{lost hot}} = m_{\text{hot}} \times c \times \Delta T \][/tex]
[tex]\[ Q_{\text{lost hot}} = 1000 \, \text{g} \times 4.18 \, \text{J/g°C} \times 65.33 \, \text{°C} \][/tex]
Calculate [tex]\( Q_{\text{lost hot}} \)[/tex]:
[tex]\[ Q_{\text{lost hot}} = 1000 \times 4.18 \times 65.33 \][/tex]
[tex]\[ Q_{\text{lost hot}} = 273079.4 \, \text{J} \][/tex]
Therefore, the energy lost by the hot water is approximately 273079.4 Joules.
