How long does it take a 1000W electric kettle to bring 1.0L of water to the boiling point if the initial temperature is 15oc and the kettle is made of 400g of iron? Assume that no heat was lost to the surroundings and that the kettle was 100% efficient?
1 Liter of Water = 1000 grams
ΔT = (100-15)° = 85°
Specific heat capacity of water = 4.184 J/gC°
Specific heat capacity of iron = 0.450 J/gC°
there is an amount of energy required to get 1L of water to 100°C and likewise an energy for the 400g of iron.
energy spent is measured Joules.
Joule = Watt*sec
Joules = (1000W)(T sec) = (4.184 J/gC°)(1000g)(85°) + (0.450 J/gC°)(400g)(85°)
T sec = [ 355640J + 15300J ]/(1000 W)
= 370.94 sec
To find out how long it takes for the electric kettle to bring the water to the boiling point, we need to calculate the amount of heat required to raise the temperature of the water and the kettle from 15°C to 100°C.
To calculate the heat required to raise the temperature of the water, we use the specific heat capacity of water, which is about 4.18 J/g°C.
The mass of the water is 1.0L, which is equivalent to 1000g. So, the heat required to raise the temperature of the water can be calculated as follows:
Heat (Q) = mass (m) x specific heat capacity (c) x temperature change (ΔT)
= 1000g x 4.18 J/g°C x (100°C - 15°C)
= 1000g x 4.18 J/g°C x 85°C
Next, to calculate the heat required to raise the temperature of the kettle, we use the specific heat capacity of iron, which is about 0.45 J/g°C.
The mass of the kettle is given as 400g. So, the heat required to raise the temperature of the kettle can be calculated as follows:
Heat (Q) = mass (m) x specific heat capacity (c) x temperature change (ΔT)
= 400g x 0.45 J/g°C x (100°C - 15°C)
= 400g x 0.45 J/g°C x 85°C
Now that we have the total heat required to raise the temperature of both the water and the kettle, we can calculate the time required using the power of the kettle, which is 1000W.
Time (t) = Heat (Q) / Power (P)
For this calculation, we need to convert the mass of the water and kettle to kilograms and convert the units of heat from Joules (J) to kilojoules (kJ) since power is given in Watts (W).
So, the total time required can be calculated as:
Time (t) = (Heat water + Heat kettle) / Power
= ((1000g x 4.18 J/g°C x 85°C) + (400g x 0.45 J/g°C x 85°C)) / 1000W
= ((1000g x 4.18 J/g°C x 85°C) + (400g x 0.45 J/g°C x 85°C)) / 1000 J/W
= ((418000 J + 15300 J) / 1000 J) / 1000 W
= 433300 J / 1000 J / 1000 W
= 0.4333 s
Therefore, it would take approximately 0.4333 seconds for the 1000W electric kettle to bring 1.0L of water to the boiling point if no heat is lost to the surroundings and the kettle is 100% efficient.
To determine how long it takes for the electric kettle to bring 1.0L of water to the boiling point, we need to calculate the amount of heat energy required and then calculate the time based on the power of the kettle.
Step 1: Calculate the amount of heat energy required.
The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change required to heat the water from 15°C to 100°C is 100 - 15 = 85°C.
The mass of water is 1.0 kg (since 1.0L of water is equivalent to 1.0 kg).
So, the amount of heat energy required can be calculated as follows:
Q = mass × specific heat capacity × temperature change
Q = 1000 g × 4.18 J/g°C × 85°C
Q = 356,300 J or 356.3 kJ
Step 2: Calculate the time required.
The power of the electric kettle is given as 1000W.
The relationship between power, energy, and time is P = E/t, where P is power, E is energy, and t is time.
Rearranging the formula, we can solve for time:
t = E/P
t = 356.3 kJ / 1000 W
t = 0.3563 hours or 21.38 minutes
Therefore, it would take approximately 21.38 minutes for the 1000W electric kettle to bring 1.0L of water to the boiling point.