Respuesta :
The heat that is required to convert 422 g of liquid H₂O at 23.5 °c into steam at 150 °c can be calculated as follows:
The heat capacity of water = 4184 J K⁻¹ C⁻¹
Therefore, heat required to warm the water from
23.5 °C to 100.0 °C.
= [tex]m\times c\times \Delta T[/tex]
Here,
m=0.422 kg
c=4184 J K⁻¹ C⁻¹
[tex]\Delta T[/tex] =100-23.5
so, heat required to warm the water from 23.5 °C to 100.0 °C
= 0.422 × 4184 × (100-23.5)
= 135072.072 J
Latent heat of vaporization of water is 2260 kJ/kg
Thus the heat will be 0.422 × 2260000 = 953720 J
Heat required to raise steam from 100 to 150
2000 × 0.422 ×50 = 42200 J
Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules
Heat transfer from the system changes the phase of the substance into another state when temperature and pressure are applied. The heat required for the conversion of the water into steam is 1330992.07 J.
What is heat required by the water?
The temperature and the heat transfer includes the mass of the substance and the change in the temperature.
Given,
- Mass of the water (m)= 0.422 kg
- Heat capacity of the water (c) = [tex]4184 \rm \;J K^{-1} C^{-1}[/tex]
- Change in temperature [tex]\rm (\Delta T)[/tex] = 100-23.5
The formula to calculate the heat required is:
[tex]\rm Q = mc \Delta T[/tex]
Substituting values in the equation:
[tex]\begin{aligned}&= 0.422 \times 4184 \times (100-23.5)\\\\&= 135072.072 \;\rm J\end{aligned}[/tex]
We know that the latent heat of vaporization is 2260 kJ/kg. So, the heat will be:
[tex]\rm 0.422\times 2260000 = 953720 \;\rm J[/tex]
The heat that will be needed to raise steam is:
[tex]2000 \times 0.422 \times 50 = 42200 \rm \;J[/tex]
The total heat needed is:
[tex]135072.072 + 953720 + 42200 = 1330992.07 \;\rm Joules[/tex]
Therefore, 1330992.07 J of heat is required for the conversion of the water into steam.
Learn more about the heat here:
https://brainly.com/question/26107547
