An electric heater rated at 1.0\text{ kW} is immersed in 0.20\text{ kg} of water at its bo… — JAMB UTME

An electric heater rated at $1.0\text{ kW}$ is immersed in $0.20\text{ kg}$ of water at its boiling point of $100^\circ\text{C}$ . Assuming all the energy supplied by the heater is used to vaporize the water and there is no heat loss to the surroundings, how long will it take for all the water to completely turn into steam? (Take the specific latent heat of vaporization of water $L_v = 2.25 \times 10^6\text{ J kg}^{-1}$ )

$450\text{ s}$ Cevap
B
$45,000\text{ s}$
C
$84\text{ s}$
D
$2250\text{ s}$

Cevap

The total time required to completely vaporize the water is

450\text{ s}

At the boiling point of

100^\circ\text{C}

, all thermal energy supplied goes into changing the liquid water to steam without increasing its temperature. The required heat is

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 450,000\text{ J}

. Dividing energy by the heater power (

1000\text{ W}

) yields

450\text{ s}

Adım Adım Çözüm

Calculate the total thermal energy required to vaporize the water at constant temperature.

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 4.5 \times 10^5\text{ J} = 450,000\text{ J}

Phase change at the boiling point occurs at constant temperature, so only latent heat is required.

Convert the power of the heater from kilowatts to watts.

P = 1.0\text{ kW} = 1000\text{ W} = 1000\text{ J s}^{-1}

Standard SI unit for electrical power is watts (joules per second).

Determine the time required using the relationship between energy, power, and time.

t = \frac{Q}{P} = \frac{450,000\text{ J}}{1000\text{ W}} = 450\text{ s}

Power is the rate of energy transfer (

P = \frac{Q}{t}

Anahtar Kavram

Latent heat of vaporization and electrical power heat transfer at constant temperature during a phase change.

Tahmini Süre:1m 30s

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