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Difficulty: Very hardLatent Heat and Changes of State

An electric heater rated at 690 W690\text{ W} is immersed in a thermally insulated container holding a mixture of 0.50 kg0.50\text{ kg} of ice and 0.50 kg0.50\text{ kg} of liquid water in equilibrium at 0∘C0^\circ\text{C}. The heater is operated for 5.0 minutes5.0\text{ minutes}. Assuming negligible heat capacity for the container and no heat loss to the surroundings, what is the final equilibrium temperature of the mixture?
(Specific latent heat of fusion of ice Lf=3.3×105 J kg−1L_f = 3.3 \times 10^5\text{ J kg}^{-1}, specific heat capacity of water cw=4200 J kg−1 K−1c_w = 4200\text{ J kg}^{-1}\text{ K}^{-1})

  1. A
    0∘C0^\circ\text{C}
  2. 10∘C10^\circ\text{C}Answer
  3. C
    20∘C20^\circ\text{C}
  4. D
    5∘C5^\circ\text{C}

Answer

The final equilibrium temperature of the mixture is 10∘C10^\circ\text{C}.
The total energy supplied by the 690 W690\text{ W} heater in 300 s300\text{ s} is 207,000 J207,000\text{ J}. Melting all 0.50 kg0.50\text{ kg} of ice at 0∘C0^\circ\text{C} requires 165,000 J165,000\text{ J}. The remaining 42,000 J42,000\text{ J} heats the combined 1.00 kg1.00\text{ kg} of water (initial water + melted ice) through ΔT=42,0001.00×4200=10∘C\Delta T = \frac{42,000}{1.00 \times 4200} = 10^\circ\text{C}, reaching a final equilibrium temperature of 10∘C10^\circ\text{C}.

Step-by-Step Solution

1
Calculate the total thermal energy provided by the electric heater.
Qtotal=P×t=690 W×(5.0×60 s)=207,000 JQ_{\text{total}} = P \times t = 690\text{ W} \times (5.0 \times 60\text{ s}) = 207,000\text{ J}.
Power multiplied by time yields total energy supplied.
2
Determine the energy required to completely melt the 0.50 kg0.50\text{ kg} of ice at 0∘C0^\circ\text{C}.
Qmelt=mice×Lf=0.50 kg×330,000 J kg−1=165,000 JQ_{\text{melt}} = m_{\text{ice}} \times L_f = 0.50\text{ kg} \times 330,000\text{ J kg}^{-1} = 165,000\text{ J}.
Latent heat of fusion changes state from solid ice to liquid water at constant temperature 0∘C0^\circ\text{C}.
3
Calculate the remaining thermal energy available to increase the temperature.
Qrem=207,000 J−165,000 J=42,000 JQ_{\text{rem}} = 207,000\text{ J} - 165,000\text{ J} = 42,000\text{ J}.
After complete melting, excess energy goes into sensible heating.
4
Calculate the total mass of liquid water and the resulting temperature rise.
mtotal=0.50 kg (melted ice)+0.50 kg (initial water)=1.00 kgm_{\text{total}} = 0.50\text{ kg (melted ice)} + 0.50\text{ kg (initial water)} = 1.00\text{ kg}. ΔT=Qremmtotal×cw=42,000 J1.00 kg×4200 J kg−1 K−1=10∘C\Delta T = \frac{Q_{\text{rem}}}{m_{\text{total}} \times c_w} = \frac{42,000\text{ J}}{1.00\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1}} = 10^\circ\text{C}.
All water now absorbs energy to raise the temperature.

Key Concept

Phase changes occur at constant temperature (latent heat Q=mLQ = mL). Once the phase change is complete, additional thermal energy increases temperature as sensible heat (Q=mcΔTQ = mc\Delta T) using the total combined mass.
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Topics

Latent Heat and Changes of State

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