An electric heater rated at 500\text{ W} is used to melt 0.20\text{ kg} of ice initially a… — JAMB UTME

An electric heater rated at $500\text{ W}$ is used to melt $0.20\text{ kg}$ of ice initially at $0^\circ\text{C}$ and heat the resulting water to $20^\circ\text{C}$ . Assuming no heat losses to the surroundings, what is the time required in seconds? (Specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4.2 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}$ )

Cevap: 168 s

Cevap

168 s

The total energy required is the sum of the heat required for melting (

mL_f = 67,200\text{ J}

) and the heat required for warming the liquid (

mc\Delta T = 16,800\text{ J}

), giving

84,000\text{ J}

. Dividing this by the heater power rating of

500\text{ W}

gives

168\text{ s}

Adım Adım Çözüm

Calculate thermal energy needed to change phase from ice to water at

0^\circ\text{C}

Q_1 = m L_f = 0.20\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 67,200\text{ J}

Latent heat of fusion accounts for phase change at constant temperature.

Calculate thermal energy needed to raise water temperature from

0^\circ\text{C}

20^\circ\text{C}

Q_2 = m c \Delta T = 0.20\text{ kg} \times 4.2 \times 10^3\text{ J kg}^{-1}\text{K}^{-1} \times 20\text{ K} = 16,800\text{ J}

Sensible heat increases kinetic energy of liquid water molecules.

Sum energy required for both phase change and temperature change

Q_{\text{total}} = Q_1 + Q_2 = 67,200\text{ J} + 16,800\text{ J} = 84,000\text{ J}

Total energy is the sum of sensible heat and latent heat.

Calculate heating time using heater power rating

t = \frac{Q_{\text{total}}}{P} = \frac{84,000\text{ J}}{500\text{ W}} = 168\text{ s}

Power is defined as energy transferred per unit time (

P = \frac{Q}{t}

Anahtar Kavram

Energy balance combining latent heat and sensible heat

Tahmini Süre:2m 0s

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