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Difficulty: Very hardLatent Heat and Changes of State

An electric heater rated at 1.0 kW1.0\text{ kW} is used to convert 0.50 kg0.50\text{ kg} of ice initially at −10∘C-10^\circ\text{C} completely into water at 50∘C50^\circ\text{C}. Assuming zero thermal energy loss to the surroundings, what is the total time required for this conversion?

(Take specific heat capacity of ice = 2100 J kg−1 K−12100\text{ J kg}^{-1}\text{ K}^{-1}, specific latent heat of fusion of ice = 3.36×105 J kg−13.36 \times 10^5\text{ J kg}^{-1}, specific heat capacity of water = 4200 J kg−1 K−14200\text{ J kg}^{-1}\text{ K}^{-1})

  1. 283.5 s283.5\text{ s}Answer
  2. B
    273.0 s273.0\text{ s}
  3. C
    1953.5 s1953.5\text{ s}
  4. D
    126.0 s126.0\text{ s}

Answer

283.5 s283.5\text{ s}
The correct response of 283.5 s283.5\text{ s} accurately accounts for all three distinct phases of thermal absorption: raising the temperature of solid ice from −10∘C-10^\circ\text{C} to 0∘C0^\circ\text{C} (10,500 J10,500\text{ J}), melting the ice to water at constant temperature (168,000 J168,000\text{ J}), and raising the liquid water temperature to 50∘C50^\circ\text{C} (105,000 J105,000\text{ J}). Dividing the total energy of 283,500 J283,500\text{ J} by the heater power of 1000 W1000\text{ W} yields 283.5 s283.5\text{ s}.

Step-by-Step Solution

1
Calculate the heat required to raise the temperature of ice from −10∘C-10^\circ\text{C} to 0∘C0^\circ\text{C} (Q1Q_1)
Q1=m⋅cice⋅ΔT1=0.50×2100×(0−(−10))=10,500 JQ_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1 = 0.50 \times 2100 \times (0 - (-10)) = 10,500\text{ J}
Ice must reach its melting point at 0∘C0^\circ\text{C} before any phase change can occur.
2
Calculate the latent heat required to melt ice at 0∘C0^\circ\text{C} into water at 0∘C0^\circ\text{C} (Q2Q_2)
Q2=m⋅Lf=0.50×3.36×105=168,000 JQ_2 = m \cdot L_f = 0.50 \times 3.36 \times 10^5 = 168,000\text{ J}
Phase change occurs at a constant temperature of 0∘C0^\circ\text{C} using latent heat of fusion.
3
Calculate the heat required to raise the temperature of the resulting water from 0∘C0^\circ\text{C} to 50∘C50^\circ\text{C} (Q3Q_3)
Q3=m⋅cwater⋅ΔT2=0.50×4200×(50−0)=105,000 JQ_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 = 0.50 \times 4200 \times (50 - 0) = 105,000\text{ J}
Once completely melted, sensible heat is absorbed by liquid water up to the target temperature.
4
Calculate total heat energy required (QtotalQ_{\text{total}}) and convert power to watts
Qtotal=10,500+168,000+105,000=283,500 JQ_{\text{total}} = 10,500 + 168,000 + 105,000 = 283,500\text{ J}, and P=1.0 kW=1000 WP = 1.0\text{ kW} = 1000\text{ W}
Total energy is the sum of all individual stage energies.
5
Determine the time required (tt)
t=QtotalP=283,5001000=283.5 st = \frac{Q_{\text{total}}}{P} = \frac{283,500}{1000} = 283.5\text{ s}
Power is defined as energy per unit time (P=QtP = \frac{Q}{t}).

Key Concept

Multi-stage thermal energy balance combining sensible heat (Q=mcΔTQ = m c \Delta T) and latent heat of fusion (Q=mLfQ = m L_f).

Alternative Method

Calculate energy per unit mass first: qtotal=ciceΔT1+Lf+cwaterΔT2=(2100×10)+336000+(4200×50)=567,000 J kg−1q_{\text{total}} = c_{\text{ice}}\Delta T_1 + L_f + c_{\text{water}}\Delta T_2 = (2100 \times 10) + 336000 + (4200 \times 50) = 567,000\text{ J kg}^{-1}. Then total energy Q=0.50×567,000=283,500 JQ = 0.50 \times 567,000 = 283,500\text{ J}, leading to t=283,5001000=283.5 st = \frac{283,500}{1000} = 283.5\text{ s}.
Estimated Time:3m 0s
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Latent Heat and Changes of State

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