To calculate the amount of heat energy transferred during the cooling of water, we can use the formula:
\[ Q = mc\Delta T \]
where:
- \( Q \) = heat energy transferred (in joules)
- \( m \) = mass of the substance (in grams)
- \( c \) = specific heat capacity (in J/g°C)
- \( \Delta T \) = change in temperature (in °C)
Given:
- Mass of water: \( m = 5.0 , \text{kg} = 5000 , \text{g} \) (since 1 kg = 1000 g)
- Specific heat capacity: \( c = 4.18 , \text{J/g°C} \)
- Initial temperature: \( T_i = 80°C \)
- Final temperature: \( T_f = 20°C \)
- Change in temperature: \( \Delta T = T_f - T_i = 20°C - 80°C = -60°C \)
Now substituting the values into the formula:
\[ Q = (5000 , \text{g}) \times (4.18 , \text{J/g°C}) \times (-60 , \text{°C}) \]
Calculating:
\[ Q = 5000 \times 4.18 \times -60 \]
\[ Q = 5000 \times 4.18 \times -60 = 5000 \times -250.8 = -1254000 , \text{J} \]
Since heat energy transferred out of the system is typically reported as a positive value, we can say:
\[ Q = 1254000 , \text{J} \]
Thus, the amount of heat energy transferred when 5.0 kg of water is cooled from 80°C to 20°C is 1,254,000 J or 1.25 MJ.