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Auto Principle AENG. Instructions: Answer Any Five Questions. Q.1a. Define/Explain the following heat related terms: i. Sensible Heat ii. Latent Heat iii. Latent Heat of fusion iv. Latent Heat of Vaporization.

Physics

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To calculate the total heat required to evaporate the water, we need to consider two parts:

  1. The sensible heat required to raise the temperature of the water from 15∘C15^\circ\text{C} to its boiling point (100∘C100^\circ\text{C}).
  2. The latent heat required to change the water from liquid to steam at its boiling point.

Given values:
Volume of water (VV) = 15 litres15 \text{ litres}
Mass of water (mm) = 15 kg15 \text{ kg} (since 1 litre1 \text{ litre} of water has a mass of 1 kg1 \text{ kg})
Initial temperature (T1T_1) = 15∘C15^\circ\text{C}
Boiling temperature (T2T_2) = 100∘C100^\circ\text{C}
Specific Heat Capacity of water (cc) = 4.18 kJ/kg∘C4.18 \text{ kJ/kg}^\circ\text{C}
Latent Heat of vaporization of steam (LvL_v) = 2256 kJ/kg2256 \text{ kJ/kg}

  1. Calculate the change in temperature (ΔT\Delta T).

    ΔT=T2−T1=100∘C−15∘C=85∘C\Delta T = T_2 - T_1 = 100^\circ\text{C} - 15^\circ\text{C} = 85^\circ\text{C}
  2. Calculate the sensible heat (QsQ_s) required to raise the water temperature.

    The formula for sensible heat is Qs=mcΔTQ_s = mc\Delta T.

    Qs=(15 kg)×(4.18 kJ/kg∘C)×(85∘C)Q_s = (15 \text{ kg}) \times (4.18 \text{ kJ/kg}^\circ\text{C}) \times (85^\circ\text{C}) Qs=6300.9 kJQ_s = 6300.9 \text{ kJ}
  3. Calculate the latent heat (QLQ_L) required to evaporate the water.

    The formula for latent heat of vaporization is QL=mLvQ_L = mL_v.

    QL=(15 kg)×(2256 kJ/kg)Q_L = (15 \text{ kg}) \times (2256 \text{ kJ/kg}) QL=33840 kJQ_L = 33840 \text{ kJ}
  4. Calculate the total heat (QtotalQ_{total}) required.

    Qtotal=Qs+QLQ_{total} = Q_s + Q_L Qtotal=6300.9 kJ+33840 kJQ_{total} = 6300.9 \text{ kJ} + 33840 \text{ kJ} Qtotal=40140.9 kJQ_{total} = 40140.9 \text{ kJ}

    The total heat required to evaporate all the water is 40140.9 kJ\boxed{\text{40140.9 kJ}}.

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