Answer:
48.37514 kj
Explanation:
Given data:
Mass of water = 163 g
Initial temperature = 29°C
Final temperature = 100°C
Heat added = ?
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 of water is 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 = Final temperature - initial temperature
ΔT = 100°C - 29°C
ΔT = 71°C
Q = 163 g × 4.18 j/g.°C × 71°C
Q = 48375.14 j
Joule to Kj conversion:
48375.14 /1000 = 48.37514 kj
