An aluminum kettle weighing 0.8 kg contains 3 liters of water at 20 ° C.

An aluminum kettle weighing 0.8 kg contains 3 liters of water at 20 ° C. Determine the amount of heat that needs to be spent on heating the kettle with water to a boil.

To find the amount of heat Q = Qal + Qw
Qal = m * c * (t2-t1) -the amount of heat for heating aluminum
mal = 0.8 kg – weight of aluminum
Sal = 920 J / kg * C-specific heat of aluminum
t2 = 100C-boiling point of water
Qal = 0.8 * 920 * (100-20) = 58880 J
Qv = m * c * (t2-t1)
Cw = 4200J / kg * s
m = p * V
p = 1000kg / m ^ 3-density of water
V = 3 l = 0.003 m ^ 3 – volume of water
m = 0.003 * 1000 = 3kg
Qv = 3 * 4200 * (100-20) = 1008000 J
Q = 1008000 + 58880 = 1066880 J