Answer:
Q = 1379.4 J
Explanation:
Given data:
Mass of water = 22 g
Initial temperature = 18°C
Final temperature = 33°C
Heat absorbed = ?
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 = 33°C - 18 °C
ΔT = 15°C
Q = 522 g ×4.18 J/g.°C× 15°C
Q = 1379.4 J
