To tackle this problem, we need to break it down into two parts: first, calculating the time required to heat the water to its boiling point, and second, determining how long it takes to boil away enough water to expose the kettle's heating element. Let's dive into the calculations step by step.
Part A: Heating Water to Boiling Point
We start with the first part where we need to find out how long it takes to heat 1.6 kg of water from 20°C to 100°C using a 3 kW kettle. The specific heat capacity of water is approximately 4,186 J/(kg·°C).
Step 1: Calculate the Temperature Change
The temperature change (ΔT) can be calculated as follows:
- Initial temperature (T_initial) = 20°C
- Final temperature (T_final) = 100°C
- ΔT = T_final - T_initial = 100°C - 20°C = 80°C
Step 2: Calculate the Energy Required
Next, we calculate the energy (Q) required to heat the water using the formula:
Q = m × c × ΔT
- m = mass of water = 1.6 kg
- c = specific heat capacity of water = 4,186 J/(kg·°C)
- ΔT = 80°C
Plugging in the values:
Q = 1.6 kg × 4,186 J/(kg·°C) × 80°C
Q = 1.6 × 4,186 × 80 = 534,784 J
Step 3: Calculate the Time Required
Now, we need to find the time (t) it takes to provide this amount of energy using the kettle's power:
Power (P) = 3 kW = 3,000 W
Using the formula:
t = Q / P
Substituting the values:
t = 534,784 J / 3,000 W
t ≈ 178.26 seconds
Part B: Boiling Away Water to Expose the Heating Element
Now, let's move on to the second part of the question. We need to find out how long it takes to boil away enough water to expose the heating element, given that the kettle requires 0.5 kg of water to cover it.
Step 1: Determine the Mass of Water to be Boiled Away
The total mass of water we have is 1.6 kg, and we need to keep 0.5 kg to cover the heating element. Therefore, the mass of water that needs to be boiled away is:
Mass to boil away = Total mass - Mass to cover element
Mass to boil away = 1.6 kg - 0.5 kg = 1.1 kg
Step 2: Calculate the Energy Required to Boil the Water
To boil water, we need to consider the latent heat of vaporization, which is approximately 2,260,000 J/kg for water. The energy required (Q_boil) to boil 1.1 kg of water is:
Q_boil = m × L
- m = 1.1 kg
- L = latent heat of vaporization = 2,260,000 J/kg
Thus:
Q_boil = 1.1 kg × 2,260,000 J/kg = 2,486,000 J
Step 3: Calculate the Time Required to Boil the Water
Using the kettle's power again, we can find the time required to boil away this amount of water:
t_boil = Q_boil / P
Substituting the values:
t_boil = 2,486,000 J / 3,000 W
t_boil ≈ 828.67 seconds
Summary of Results
To summarize:
- The least possible time to bring the water to a boil is approximately 178.26 seconds.
- The minimum time to boil away enough water to expose the heating element is approximately 828.67 seconds.
These calculations illustrate the principles of thermodynamics and energy transfer, showing how power, mass, and temperature changes interact in practical scenarios like boiling water in a kettle.
