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Difficulty: EasyLatent Heat and Changes of State

What quantity of heat energy is required to completely melt 0.50 kg0.50\text{ kg} of ice at 0∘C0^\circ\text{C} into water at 0∘C0^\circ\text{C}? (Specific latent heat of fusion of ice = 3.3×105 J kg−13.3 \times 10^5\text{ J kg}^{-1})

  1. 1.65×105 J1.65 \times 10^5\text{ J}Answer
  2. B
    3.30×105 J3.30 \times 10^5\text{ J}
  3. C
    6.60×105 J6.60 \times 10^5\text{ J}
  4. D
    1.65×107 J1.65 \times 10^7\text{ J}

Answer

The quantity of heat energy required is 1.65×105 J1.65 \times 10^5\text{ J}.
The quantity of heat required for melting at constant temperature depends only on mass and specific latent heat of fusion: Q=mLf=0.50×3.3×105=1.65×105 JQ = mL_f = 0.50 \times 3.3 \times 10^5 = 1.65 \times 10^5\text{ J}.

Step-by-Step Solution

1
Identify the given values and formula for change of state
Mass m=0.50 kgm = 0.50\text{ kg}, Specific latent heat of fusion Lf=3.3×105 J kg−1L_f = 3.3 \times 10^5\text{ J kg}^{-1}. The formula is Q=mLfQ = m L_f.
During a phase change at constant temperature, latent heat is absorbed without a change in temperature.
2
Calculate total heat energy QQ
Q=0.50 kg×3.3×105 J kg−1=1.65×105 JQ = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J kg}^{-1} = 1.65 \times 10^5\text{ J}.
Multiplying mass by the specific latent heat gives the total energy transferred.

Key Concept

Latent Heat of Fusion
Estimated Time:45s
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Topics

Latent Heat and Changes of State

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