A.
The energy of the hot water is 482630400 J
Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,
c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C
Substituting the values of the variables into the equation, we have
Q = mcΔT
Q = ρVcΔT
Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C
Q = 482630400 J
So, the energy of the hot water is 482630400 J
B.
The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.
Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m
Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J
So, ΔU = mgΔh
ΔU = mg(h - h')
making h subject of the formula, we have
h = h' + ΔU/mg
Substituting the values of the variables into the equation, we have
h = h' + ΔU/mg
h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)
h = 0 m + 482630400 J/(9800 kgm/s²)
h = 0 m + 49248 m
h = 49248 m
So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.
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