Answer:
8.99 Kcal
Explanation:
Given data:
Mass of water = 100 g
Initial temperature = 10°C
Final temperature = 100°C
Calories take = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity = 4.18 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 100°C - 10°C
ΔT = 90 °C
Q = 100 g × 4.18 J/g.°C × 90 °C
Q = 37620 J
Joule to calories:
37620 J × 1 kcal / 4184 J
8.99 Kcal
