Answer:
Q = 3877 KJ
Explanation:
Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:
Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)
Q = Q₁ + Q₂ ----- equation (1)
Now, for Q₁:
Q₁ = m C ΔT
where,
m = Mass of Water = 1.5 kg
C = Specific Heat of Water = 4187 J/kg.°C
ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C
Therefore,
Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)
Q₁ = 490 x 10³ J =490 KJ
Now, for Q₂:
Q₂ = m H
where,
m = Mass of Water = 1.5 kg
H = Heat of Vaporization of Water = 2258 KJ/kg
Therefore,
Q₂ = (1.5 kg)(2258 KJ/kg)
Q₂ = 3387 KJ
Substituting the values in equation (1), we get:
Q = Q₁ + Q₂
Q = 490 KJ + 3387 KJ
Q = 3877 KJ
