Answer:
Q = 10.10059 KJ
Step-by-step explanation:
Given data:
Amount of water = 13.0 g
Initial temperature = 52.5°C
Final temperature = 238.2°C
Heat required to convert it into steam = ?
Solution:
Specific heat capacity of water is 4.184 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 = 238.2°C - 52.5°C
ΔT = 185.7 °C
by putting values,
Q = 13.0 g × 4.184 J/g.°C × 185.7 °C
Q = 10100.59 J
Jolue to Kj:
10100.59j × 1 KJ/1000 J
Q = 10.10059 KJ