Answer:
Q = 66984.43 J
Explanation:
Given data:
Mass of water = 1.8 Kg (1.8Kg×1000 g/1 Kg =1800 g)
Initial temperature = 8.4°C
Final temperature = 17.3°C
Specific heat capacity of water = 4.1813 J/g.°C
Heat required = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 17.3°C - 8.4°C
ΔT = 8.9°C
Q = 1800 g × 4.1813 J/g.°C ×8.9°C
Q = 66984.43 J
