Pure water boils at 100 ∘ C, has a specific latent heat of vaporisation of 2260kJ/kg and a specific heat capacity of 4200J/(kg ∘ C).Part AEnergy to boil waterAll questions in this part are answered correctlyHow much energy is required for 2.0kg of water to boil away if it is already at 100 ∘ C?Value?UnitskJCorrect!Hint 1Hint 2Part BWater starting at 40 ∘ CHow much energy would be required if the water started at 40 ∘ C

Part A: Energy to boil water

The energy required to boil water can be calculated using the formula:

Q = mL

where:

• Q is the heat energy,
• m is the mass of the substance, and
• L is the specific latent heat of vaporisation.

Given:

• m = 2.0 kg
• L = 2260 kJ/kg

Substituting these values into the formula gives:

Q = 2.0 kg * 2260 kJ/kg = 4520 kJ

So, 4520 kJ of energy is required for 2.0 kg of water to boil away if it is already at 100 ∘ C.

Part B: Water starting at 40 ∘ C

If the water starts at 40 ∘ C, we first need to calculate the energy required to heat the water from 40 ∘ C to 100 ∘ C. This can be calculated using the formula:

Q = mcΔT

where:

• Q is the heat energy,
• m is the mass of the substance,
• c is the specific heat capacity, and
• ΔT is the change in temperature.

Given:

• m = 2.0 kg
• c = 4200 J/(kg ∘ C)
• ΔT = 100 ∘ C - 40 ∘ C = 60 ∘ C

Substituting these values into the formula gives:

Q = 2.0 kg * 4200 J/(kg ∘ C) * 60 ∘ C = 504000 J = 504 kJ

Adding this to the energy required to boil the water gives:

Total energy = 504 kJ + 4520 kJ = 5024 kJ

So, 5024 kJ of energy would be required if the water started at 40 ∘ C.