Answer:
1.2 × 10⁴ cal
Explanation:
Given data
- Initial temperature: 80 °C
We can calculate the heat released by the water ([tex]Q_w[/tex]) when it cools using the following expression.
[tex]Q_w = c \times m \times (T_f - T_i)[/tex]
where
c is the specific heat capacity of water (1 cal/g.°C)
[tex]Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal[/tex]
According to the law of conservation of energy, the sum of the heat released by the water ([tex]Q_w[/tex]) and the heat absorbed by the reaction ([tex]Q_r[/tex]) is zero.
[tex]Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal[/tex]
