Answer:
Q = 906.882 J
Explanation:
Given data:
Mass of water = 25.5 g
Initial temperature = 14.0 °C
Final temperature = 22.5 °C
Specific heat capacity of water = 4.184 J/g.°C
Solution:
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 = 22.5 °C - 14.0 °C
ΔT = 8.5 °C
Q = 25.5 g × 4.184 J/g.°C × 8.5 °C
Q = 906.882 J
