Answer:
a) 2*10^9 J
b) 764.8 kg
Explanation:
Given that
Energy of charge, q = 20 C
Potential difference, ΔV = 1*10^2 MV = 1*10^2 * 10^6 V = 1*10^8 V
a)
To find the energy dissipated, we use the formula
ΔU = qΔV
ΔU = 20 * 1*10^8
ΔU = 2*10^9 J
b)
Change in temperature, ΔT = 100 - 15°
ΔT = 85° C
Change in energy, ΔU = 2*10^9 J
Specific heat of water, C = 4180 j./Kg.K
Latent heat of vaporization, L(v) = 2.26*10^6 J/Kg
Q1 = mcΔT
Q2 = mL(v)
Net energy needed, U = Q1 + Q2
U = mcΔT + mL(v)
U = m (cΔT + L(v))
m = U /[cΔT + L(v)]
Being that we have all the values, we then substitute
m = 2*10^9 / [(4180 * 85) + 2.26*10^6]
m = 2*10^9 / (3.553*10^5 + 2.26*10^6]
m = 2*10^9 / 2.615*10^6
m = 764.8 kg
c)
Having 765 kg of steam at the temperature would have extreme effect on the tree, damaging it permanently. Possibly even blowing it to pieces
