Examkin

Question

Difficulty: HardLatent Heat and Changes of State

An electric immersion heater rated at 500 W500\text{ W} is used to heat a liquid of mass 0.40 kg0.40\text{ kg} from an initial temperature of 25∘C25^\circ\text{C} to its boiling point at 75∘C75^\circ\text{C}, and then vaporize half of the liquid at constant temperature. If the specific heat capacity of the liquid is 2500 J kg−1 K−12500\text{ J kg}^{-1}\text{ K}^{-1} and its specific latent heat of vaporization is 5.0×105 J kg−15.0 \times 10^5\text{ J kg}^{-1}, what is the total time required for this process?

  1. 300 s300\text{ s}Answer
  2. B
    500 s500\text{ s}
  3. C
    200 s200\text{ s}
  4. D
    10,100 s10,100\text{ s}

Answer

The total time required for the process is 300 s300\text{ s}.
The correct response combines the energy needed for thermal elevation to the boiling point (50,000 J50,000\text{ J}) with the energy required for vaporizing half the mass at constant temperature (100,000 J100,000\text{ J}), yielding a total energy of 150,000 J150,000\text{ J}. Dividing this total energy by the 500 W500\text{ W} power rating yields exactly 300 s300\text{ s}.

Step-by-Step Solution

1
Calculate the sensible heat (Q1Q_1) required to heat 0.40 kg0.40\text{ kg} of liquid from 25∘C25^\circ\text{C} to 75∘C75^\circ\text{C}.
Q1=mcΔT=0.40 kg×2500 J kg−1 K−1×(75−25) K=50,000 JQ_1 = m c \Delta T = 0.40\text{ kg} \times 2500\text{ J kg}^{-1}\text{ K}^{-1} \times (75 - 25)\text{ K} = 50,000\text{ J}.
Before phase change can occur, the liquid must first reach its boiling point.
2
Calculate the latent heat (Q2Q_2) required to vaporize half of the liquid mass (mvap=0.20 kgm_{\text{vap}} = 0.20\text{ kg}).
Q2=mvapLv=0.20 kg×5.0×105 J kg−1=100,000 JQ_2 = m_{\text{vap}} L_v = 0.20\text{ kg} \times 5.0 \times 10^5\text{ J kg}^{-1} = 100,000\text{ J}.
Phase change at the boiling point occurs at constant temperature and depends strictly on latent heat.
3
Sum the total heat energy (QtotalQ_{\text{total}}) and compute time (tt) using heater power (P=500 WP = 500\text{ W}).
Qtotal=Q1+Q2=150,000 JQ_{\text{total}} = Q_1 + Q_2 = 150,000\text{ J}, so t=QtotalP=150,000 J500 W=300 st = \frac{Q_{\text{total}}}{P} = \frac{150,000\text{ J}}{500\text{ W}} = 300\text{ s}.
Thermal power is defined as energy supplied per unit time (P=QtP = \frac{Q}{t}).

Key Concept

Latent Heat and Energy Balance in Phase Transitions
Estimated Time:2m 30s
Rate this question

Topics

Latent Heat and Changes of State

More questions from this topic