Answer: 8400 J
Explanation:
The formula referenced in the question is:
[tex]Q=m. c. \Delta T[/tex]
Where:
[tex]Q[/tex] is the thermal energy
[tex]m=100g \frac{1 kg}{1000 g}=0.1 kg[/tex] is the mass of the water sample
[tex]c=4200 \frac{J}{kg\°C}[/tex] is the specific heat capacity of water
[tex]\Delta T=20\°C[/tex] is the variation in temperature
Solving:
[tex]Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)[/tex]
[tex]Q=8400 J[/tex] This is the thermal energy released
