Answer:
Q = 5036.9 calories
Explanation:
Given that,
Mass, m = 167.9 g
The temperature raises from 25°C to 55°C.
The specific heat of water,c = 4.184 J/g °C
We need to find the heat added to water. We know that,
[tex]Q=mc\Delta T\\\\=167.9 \times 4.184\times (55-25)\\\\=21074.8\ J[/tex]
or
Q = 5036.9 calories
So, 5036.9 calories of heat is added.
