Answer:
Q = 1698.51 J
Explanation:
Given that,
Mass of water, m = 63.4 grams
It changes temperature from 22.3C to 28.7C.
We need to find the heat involved. The heat involved due to the change in temperature is given by :
[tex]Q=mc\Delta T[/tex]
Where
c is the specific heat of water, c = 4.186 joule/gram °C
So,
[tex]Q=63.4 \times 4.186 \times (28.7- 22.3)\\\\Q=1698.51\ J[/tex]
So, 1698.51 J of heat is involved.
