20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal;
= 20.6kCal = 20600Cal
learn more:
Specific heat brainly.com/question/3032746
#learnwithBrainly
This is true i think if that is a question
- formula for density is mass divided by volume
therefore density of butter = 10.0g divided by 11.6ml = 0.8620689 g/cm³ ≈ 0.862 g/cm³ (3sf)
Atoms are (stable) if it has (valence) electrons in its outer most energy level.