Answer:
Q = 2627.25 J
Explanation:
Given that,
The mass of water, m = 28.7 g
The temperature of the water to increase from 11.3 °C to 33.2 °C
The specific heat of water is 4.18 J/gºC
The heat absorbed by the water is given by :
[tex]Q=mc\Delta T\\\\Q=28.7 \times 4.18 \times (33.2-11.3)\\\\Q=2627.25\ J[/tex]
So, 2627.25 J of heat is absorbed by the water.