The total energy required to get 1 kg of water from a temperature of 90 $\circ$ C to 120 $\circ$ C is (latent heat of vaporization is 540 kcal/kg for water;and specific heats are 1.0 kcal/(kg C)for water,and 0.48 kcal/(kg C)for steam).

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To calculate the total energy required, we need to consider the energy to heat the water from 90°C to 100°C, the energy to convert the water at 100°C to steam at 100°C (latent heat of vaporization), and the energy to heat the steam from 100°C to 120°C.1. Heating water from 90°C to 100°C:$$ \Delta Q_1 = m \cdot c_{water} \cdot \Delta T = 1 \, \text{kg} \cdot 1.0 \, \text{kcal/kg}^\circ C \cdot (100°C - 90°C) = 10 \, \text{kcal} $$2. Converting water at 100°C to steam at 100°C:$$ \Delta Q_2 = m \cdot L = 1 \, \text{kg} \cdot 540 \, \text{kcal/kg} = 540 \, \text{kcal} $$3. Heating steam from 100°C to 120°C:$$ \Delta Q_3 = m \cdot c_{steam} \cdot \Delta T = 1 \, \text{kg} \cdot 0.48 \, \text{kcal/kg}^\circ C \cdot (120°C - 100°C) = 9.6 \, \text{kcal} $$Total energy required:$$ \Delta Q_{total} = \Delta Q_1 + \Delta Q_2 + \Delta Q_3 = 10 \, \text{kcal} + 540 \, \text{kcal} + 9.6 \, \text{kcal} = 559.6 \, \text{kcal} $$

The Total Energy Required to Get 1 Kg of Water ∘\circ∘

The Total Energy Required to Get 1 Kg of Water $\circ$