Answer:
The energy needed is 45.7659 kJ
Explanation:
The given mass of the ice, m = 15 g
The initial temperature of the ice, T₁ = -4°C
The temperature
The final temperature, T₂ = 115°C
The specific heat capacity of ice, c₁ = 2.09 J/(g·°C)
The latent heat of ice, l₁ = 334 J/g
The heat capacity of water, c₂ = 4.184 J/(g·°C)
The latent heat of vaporization, l₂ = 2260 J/g
The specific heat of steam, c₃ = 2.02 J/g
Therefore, we get the energy needed, ΔQ, as follows;
ΔQ = m·c₁·(T₂ - Tₐ) + m·l₁ + m·c₂(Tₓ - Tₐ) + m·l₂ + m·c₃·(T₂ - Tₓ)
Where;
Tₐ = The melting point temperature of water = 0°C
Tₓ = The boiling point temperature of water = 100°C
∴ ΔQ = 15×2.09×(0 - (-4)) + 15 × 334 + 15×4.184×(100 - 0) + 15×2260 + 15×2.02×(115 - 100) = 45,765.9
The energy needed, ΔQ = 45,765.9 J = 45.7659 kJ.
