Question

You wish to boil 1.7 kg of water, which has a specific heat capacity of 4186 J/kg−K. The water is initially at room temperature (293 K). Water boils at 373 K. 1) How much energy must be added to the water by heating it in order for it to start boiling? You currently, have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 2) Now assume that the water is contained in a 0.3-kg aluminum pot (C Al ​ =900 J/kg−K) that is initially at 293 K just like the water. How much energy must be transferred into the pot-and-water system in order to raise its temperature to the boiling point of water?

Answer 1

To calculate the energy required to heat and boil the water, use the formula Q = mcΔT + mL, where Q is the total energy, m is the mass, c is the specific heat capacity, ΔT is the change in temperature, and L is the latent heat of vaporization. For the first question, substitute the values and simplify the equation to find the energy. For the second question, also include the specific heat capacity of the aluminum pot and add its energy contribution to the equation.

Step-by-step explanation:

1) To calculate the amount of energy required to heat and boil the water, we can use the formula:

Q = mcΔT + mL

Where:

• Q is the total energy required
• m is the mass of the water
• c is the specific heat capacity of water
• ΔT is the change in temperature
• L is the latent heat of vaporization of water

The change in temperature, ΔT = boiling temperature - initial temperature

For the first question, where there is no pot:

• m = 1.7 kg
• c = 4186 J/kg-K
• ΔT = 373 K - 293 K = 80 K
• L is the latent heat of vaporization of water, which is 2.26 x 106 J/kg

Substituting the values into the formula, we get:

Q = (1.7 kg) (4186 J/kg-K) (80 K) + (1.7 kg) (2.26 x 106 J/kg)

Simplifying the equation gives the final answer.

2) For the second question, where the water is contained in an aluminum pot, we need to take into account the specific heat capacity of the pot as well as the water. The formula remains the same, but we need to add the energy required to heat the pot as well.

• mal = 0.3 kg (mass of aluminum pot)
• cal = 900 J/kg-K (specific heat capacity of aluminum)

The change in temperature, ΔT = boiling temperature - initial temperature

Substituting the values into the formula, we get:

Q = (1.7 kg) (4186 J/kg-K) (80 K) + (1.7 kg) (2.26 x 106 J/kg) + (0.3 kg) (900 J/kg-K) (80 K)

Simplifying the equation gives the final answer.

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