How much heat energy is required to vaporize a 1.0-g ice cube at 0°c? the heat of fusion of ice is 80 cal/g.
the heat of vaporization of water is 540 cal/g, and cwater = 1.00 cal/g⋅°c.
a. 620 cal
b. 720 cal
c. 820 cal
d. 1 kcal

720 Cal

Explanation:

Quantity of heat required to melt a solid is ;

                         Q₁ = mΔ

Where

Q₁ = Quantity of heat to melt ice

m = mass of the solid

Δ = 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

Where

Q₃ = Quantity of heat to vapourize water

m = mass of the water

Δ = 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

what type of charge does it have? negative

terd muffin is the correct answer