The amount of water needed is 287 kg
Explanation:
The amount of energy that we need to produce with the power plant is
[tex]E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J[/tex]
We also know that the power plant is only 30% efficient, so the energy produced in input must be:
[tex]E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J[/tex]
The amount of water that is needed to produce this energy can be found using the equation
[tex]E_{in}=mC\Delta T[/tex]
where:
m is the amount of water
[tex]C=4186 J/kg^{\circ}C[/tex] is the specific heat capacity of water
[tex]\Delta T=10^{\circ}C[/tex] is the increase in temperature
And solving for m, we find:
[tex]m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg[/tex]
Learn more about specific heat capacity:
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