9. 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.

To solve this problem, we need to use the formula for heat transfer:

Q = m * c * ΔT

Where:
Q is the heat energy transferred (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity (in J/kg°C)
ΔT is the change in temperature (in °C)

(i) Time taken for the temperature of water to rise from 25°C to its boiling point (100°C):

1. Calculate the heat energy required to raise the temperature of the water from 25°C to 100°C.

Q = m * c * ΔT

Q = 2 kg * 4186 J/kg°C * (100°C - 25°C)

Q = 2 kg * 4186 J/kg°C * 75°C

Q = 627,900 J

2. Calculate the power of the electric kettle:

Power = 2.5 kW = 2,500 W

3. Use the relationship between power, time, and energy to find the time taken.

Power = Energy / Time

Time = Energy / Power

Time = 627,900 J / 2,500 W

Time = 251.16 seconds (approximately)

Therefore, it takes approximately 251.16 seconds for the temperature of the water to rise from 25°C to its boiling point.

(ii) Mass of water evaporated per second from the boiling water:

1. Calculate the heat energy required to change the water from boiling to steam.

Q = m * L

Where:
Q is the heat energy transferred (in Joules)
m is the mass of the substance (in kg)
L is the specific latent heat of vaporization (in J/kg)

2. The specific latent heat of vaporization (L) for water is 336,000 J/kg.

Q = m * 336,000 J/kg

3. To determine the mass of water evaporated per second, we can use the formula for power:

Power = Energy / Time

m * 336,000 J/kg = Power

m = Power / 336,000 J/kg

m = 2,500 W / 336,000 J/kg

m = 0.00744 kg/s (approximately)

Therefore, approximately 0.00744 kg (or 7.44 grams) of water is evaporated per second from the boiling water.

To calculate the time taken for the temperature of water to rise from 25 degrees Celsius to its boiling point of 100 degrees Celsius, we can use the formula:

Q = mcΔT

Where:
Q = Heat energy (in Joules)
m = Mass of the water (in kilograms)
c = Specific heat capacity of water (in J/kg·°C)
ΔT = Change in temperature (in °C)

Given:
Mass of water, m = 2 kg
Specific heat capacity of water, c is negligible (meaning it can be considered zero)
Change in temperature, ΔT = 100°C - 25°C = 75°C

(i) Calculation of time taken for the water to reach boiling point:

Since the heat capacity of the kettle is negligible, all the energy supplied will go into heating the water.

Q = mcΔT
Q = (2 kg)(0 J/kg·°C)(75°C) [As c is negligible]
Q = 0 J

Now, let's calculate the power (P) used to heat the water:

P = Q / t

Given:
Power of the kettle, P = 2.5 kW = 2500 W (since 1 kilowatt is equal to 1000 watts)

Therefore, to find the time taken, we rearrange the formula:

t = Q / P
t = 0 J / 2500 W
t = 0 seconds

So, it takes 0 seconds for the temperature of the water to rise from 25°C to 100°C.

(ii) Calculation of mass of water evaporated per second:

To calculate the mass of water evaporated per second (dm/dt), we can use the formula:

dm/dt = Q / Lv

Where:
dm/dt = Mass of water evaporated per second (in kg/s)
Q = Heat energy (in Joules)
Lv = Latent heat of vaporization of water (in J/kg)

Given:
Specific latent heat of vaporization (Lv) = 336 × 10^3 J/kg (as mentioned in the question)

Since the water reaches its boiling point, all the energy supplied will go into boiling and vaporizing the water.

Q = mcΔT
Q = (2 kg)(0 J/kg·°C)(75 °C) [As c is negligible]
Q = 0 J

Now, let's calculate the mass of water evaporated per second:

dm/dt = Q / Lv
dm/dt = 0 J / (336 × 10^3 J/kg)
dm/dt = 0 kg/s

Therefore, the mass of water evaporated per second is 0 kg/s.

Please note that these calculations assume ideal conditions and neglect any heat losses to the surroundings.