Answer :
To find out how much heat is required to convert 422 g of liquid water at 23.5 °C into steam at 150 °C, we need to consider the following steps:
Heating Water from 23.5 °C to 100 °C:
The specific heat capacity of water is [tex]c = 4.18 \text{ J/g°C}[/tex]. To calculate the heat required, use the formula:
[tex]q_1 = m \cdot c \cdot \Delta T[/tex]
Where:
[tex]m = 422 \text{ g}[/tex],[tex]\Delta T = 100 - 23.5 = 76.5 \text{ °C}[/tex]
[tex]q_1 = 422 \cdot 4.18 \cdot 76.5[/tex]
[tex]q_1 = 134,721.63 \text{ J}[/tex] or 134.7 kJ
Evaporating Water at 100 °C to Steam:
The heat of vaporization of water is [tex]L = 2260 \text{ J/g}[/tex].
[tex]q_2 = m \cdot L[/tex]
[tex]q_2 = 422 \cdot 2260[/tex]
[tex]q_2 = 953,320 \text{ J}[/tex] or 953.3 kJ
Heating Steam from 100 °C to 150 °C:
The specific heat capacity of steam is [tex]c = 2.0 \text{ J/g°C}[/tex].
[tex]q_3 = m \cdot c \cdot \Delta T[/tex]
[tex]\Delta T = 150 - 100 = 50 \text{ °C}[/tex]
[tex]q_3 = 422 \cdot 2.0 \cdot 50[/tex]
[tex]q_3 = 42,200 \text{ J}[/tex] or 42.2 kJ
Total Heat Required:
[tex]q_{\text{total}} = q_1 + q_2 + q_3[/tex]
[tex]q_{\text{total}} = 134.7 + 953.3 + 42.2[/tex]
[tex]q_{\text{total}} = 1130.2 \text{ kJ}[/tex]
The total heat required is approximately 1130.0 kJ.
Therefore, the answer is option D.
