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.