Question

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

300 grams of water cools from 80°C to 40°C?​

Answer 1

1.2 × 10⁴ cal

Step-by-step 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 * m * (T_f - T_i)`

where

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

`Q_w = (1cal)/(g.\°C) * 300g * (40\°C - 80\°C) = -1.2 * 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 * 10^(4) cal`