Question

5. How much heat (in calories) is absorbed by a reaction when

Answer 1

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

• Mass of water (m): 300 g

• Initial temperature: 80 °C

• Final temperature: 40°C

We can calculate the heat released by the water ($Q_w$) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ($Q_w$) and the heat absorbed by the reaction ($Q_r$) is zero.

$Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal$