2 kg of water is contained in a 2.5 KW electric kittle. Assuring that the heat capacity of the kettle is negligible. Calculate (i) the time taken for the temp of water to rise from 25 degree centigrade to it boiling point 100 degree centigrade. (ii) the mass of the water evaporated per second from the boiling water. S.L.H of ice = 336X10^3J/kg.
It is
Again Q = mc change T from 25 to 100.
Power (2.5kW) * time will equal this Q
Given:
Mass of water (m) = 2 kg
Initial temperature (T1) = 25°C
Final temperature (T2) = 100°C
Specific latent heat of vaporization of water (SLH) = 336 × 10^3 J/kg
Power of electric kettle (P) = 2.5 kW
To calculate (i) the time taken for the temperature of water to rise from 25°C to its boiling point of 100°C:
1. Calculate the heat energy required to raise the temperature of the water from 25°C to 100°C using the formula: Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water (c = 4,186 J/kg°C), and ΔT is the change in temperature.
Q = 2 kg × 4,186 J/kg°C × (100°C - 25°C)
Q = 2 kg × 4,186 J/kg°C × 75°C
Q = 627,900 J
2. Convert the power of the electric kettle from kilowatts to watts by multiplying by 1000 (since 1 kW = 1000 W):
P = 2.5 kW × 1000 W/kW
P = 2500 W
3. Use the formula: Q = Pt, where Q is the heat energy, P is the power, and t is the time taken.
627,900 J = 2500 W × t
t = 627,900 J / 2500 W
t ≈ 251.16 seconds
t ≈ 4.19 minutes
Therefore, the time taken for the temperature of water to rise from 25°C to its boiling point of 100°C is approximately 4.19 minutes.
To calculate (ii) the mass of water evaporated per second from the boiling water:
1. Calculate the heat energy required to convert the water from 100°C to steam using the formula: Q = mL, where Q is the heat energy, m is the mass of water evaporated, and L is the specific latent heat of vaporization of water.
Q = m × 336 × 10^3 J/kg
Since the water is boiling, its temperature remains constant at 100°C.
2. Use the formula: Q = Pt, where Q is the heat energy, P is the power, and t is the time taken. Rearrange the formula to solve for t:
t = Q / P
Since we want to find the mass evaporated per second, we need to find the mass evaporated for a time period of 1 second.
3. Substitute the given values into the formula to find the mass evaporated per second:
t = 336 × 10^3 J/kg / 2500 W
t ≈ 134.4 kg/s
Therefore, the mass of water evaporated per second from the boiling water is approximately 134.4 kg/s.
To solve this problem, we need to use the equations related to thermal energy and heating/boiling processes. Here's how you can find the answers to the two questions:
(i) The time taken for the temperature of water to rise from 25°C to its boiling point, 100°C:
First, we need to find the amount of heat energy required to raise the temperature of the water from 25°C to 100°C. The equation for the heat energy is given by:
Q = mcΔT
Where:
Q = Heat energy (in joules)
m = mass of water (in kilograms)
c = specific heat capacity of water (assumed to be 4.18 J/g°C or 4.18 kJ/kg°C)
ΔT = change in temperature (in °C)
In this case, we are given that the mass of water is 2 kg, so m = 2 kg. Also, ΔT = (100°C - 25°C) = 75°C.
Now, we can calculate the heat energy:
Q = (2 kg) * (4.18 kJ/kg°C) * (75°C)
Q = 627 kJ
Next, we need to convert the power of the electric kettle from kW to kJ/s. Since 1 kW = 1000 W and 1 J = 1 W/s, we can convert the power:
Power = 2.5 kW = 2.5 * 1000 J/s = 2500 J/s
Finally, to find the time taken (t) for the temperature to rise, we'll use the equation:
Q = Power * t
Solving for t:
t = Q / Power
t = (627 kJ) / (2500 J/s)
t = 251.6 seconds
Therefore, it would take approximately 251.6 seconds for the temperature of the water to rise from 25°C to its boiling point of 100°C.
(ii) The mass of water evaporated per second from the boiling water:
To calculate the mass of water evaporated per second, we'll use the specific latent heat (S.L.H) of water. S.L.H of ice is given as 336 × 10^3 J/kg.
The heat energy required to evaporate the mass of water (m_water) per second can be calculated using the equation:
Q = m_water * S.L.H
We know that the power of the electric kettle is 2.5 kW, so we can convert it to J/s:
Power = 2.5 kW = 2.5 * 1000 J/s = 2500 J/s
Now, we can rearrange the equation:
m_water = Q / S.L.H
Substituting the values:
m_water = (2500 J/s) / (336 × 10^3 J/kg)
m_water ≈ 0.0074405 kg/s
Therefore, approximately 0.00744 kg (or 7.44 grams) of water would be evaporated per second from the boiling water.
Note: In these calculations, we have assumed that the entire heat energy supplied by the kettle is used to heat the water and cause evaporation, neglecting any heat losses to the surroundings.