Answer :
Final answer:
To change 1.47 kg of H₂O at 68.0°C to steam at 144°C, a total of 3,495,988 J of heat is required.
Explanation:
To calculate the amount of heat required to change 1.47 kg of H₂O from 68.0°C to steam at 144°C, we need to consider two steps: heating the water to its boiling point and then vaporizing it.
First, we calculate the heat required to raise the temperature of the water from 68.0°C to its boiling point, 100°C, using the specific heat capacity of water: Q = m * C * ΔT = (1.47 kg) * (4186 J/kg/K) * (100°C - 68.0°C) = 176,698 J.
Next, we calculate the heat required to vaporize the water at 100°C using the latent heat of vaporization of water: Qv = m * Lv = (1.47 kg) * (2,257,000 J/kg) = 3,319,290 J.
Now, to calculate the total heat required, we sum up the two values: Q_total = Q + Qv = 176,698 J + 3,319,290 J = 3,495,988 J.
