Answer:
720 Cal
Explanation:
Quantity of heat required to melt a solid is ;
Q₁ = mΔ[tex]H_{fus}[/tex]
Where
Q₁ = Quantity of heat to melt ice
m = mass of the solid
Δ[tex]H_{fus}[/tex] = enthalpy of fusion (Latent heat of fusion)
Q₁ = 1 g × 80 cal/g
Q₁ = 80 cal
The amount of heat required to melt Ice is 80J
The ice is now water at 0°C. We need to calculate the amount of heat needed to boil the water from 0°C to 100°C
Q₂ = mcΔT
where;
Q₂ is quantity of heat to boil water
m is mass of water
c is specific heat capacity of water
ΔT is change in temperature
Q₂ = 1 × 1 × (100 - 0)
Q₂ = 1 × 1 × 100
Q₂ = 100 cal
We need to calculate the amount of heat needed to vapourize the water.
Q₃ = mΔ[tex]H_{vap}[/tex]
Where
Q₃ = Quantity of heat to vapourize water
m = mass of the water
Δ[tex]H_{vap}[/tex] = enthalpy of vapourization (Latent heat of vapourization)
Q₃ = 1 × 540
Q₃ = 540 cal
The total amount of heat energy that is required to vaporize a 1.0-g ice cube at 0°C
Q = Q₁ + Q₂ + Q₃
Q = 80 + 100 + 540
Q = 720 cal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
We know that when heat is supplied to a substance, change of state occurs as the object moves from a particular state of matter to another. The heat required to convert 1.0-g of ice cube at 0°C to vapor is obtained from;
H = mLfus + mcdT + mLvap
H = (1 g * 80 cal/g) + (1 g * 1.00 cal/g⋅°C * (100 - 0)) + (1 g * 540 cal/g)
H = 80 cal + 100 cal + 540 cal
H = 720 cal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
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