Commented Resolution
Work (Energy) – Power
01- The energy stored by water is the gravitational potential energy of that mass of water at a height of 120m — W=mgh=m.10.120 — W=1,200m — Power=energy/time — P o =512,000,000W — ∆t=1s — P o =W/∆t — 512,000,000=1,200m/1 — m=512,000,000/1,200 =512.106/ 12.102 — m=42.7.104 = 427,000kg + (10% of 427,000=42,700) —
m=469.700kg.
Considering the density of water as 1kg/L, we will have that a mass of 1kg occupies a volume of 1L of water — R- E
02- Power corresponds to the measure of how quickly energy is transferred — Po = W/Δt — note in this expression that the most efficient (highest power Po ) is the one that provides the greatest amount of energy (W) in the shortest amount of time (Δt), that is, it is directly proportional to the amount of energy (W) and inversely proportional to the operating time interval (Δt) — R- C
03- m=1 ton = 1,000kg — rule of three — 1,000kg – 100% — 1.5kg – p — 1000p=1.5×100 — p=150/1000 — p=0.15%
R-B
04- Since the tractor is the same, the energy it consumes can be considered constant for any type of fuel you use — lower fuel consumption per liter of a given fuel will imply a greater amount of energy released by the same liter, making the tractor run for longer — therefore, the energy released by a liter of sunflower oil will be greater than the energy released by a liter of diesel oil — A- E
05- I- Demand on Sundays is 60% of total capacity — 60% of 240MW=0.6.240 — demand on Sundays=144MW — CORRECT
II- Each turbine has maximum capacity — 240/24 = 10MW — half of them working would have 120MW — 24MH are missing to complete 144MW — 12 turbines are missing and the capacity of each one is 2MW (20% of the capacity of a turbine) — therefore, 2 x 12 = 24MW, which is needed to complete 144 — CORRECT
III- Fourteen with maximum capacity is 140MW, and 40 percent of a turbine is 4MW, therefore it will reach the 144MW we need. CORRECT
R- And
06- First, combustion occurs, which is the energy coming from a chemical process — then the paraffin falls from a certain height, causing a transformation of gravitational potential energy (E pg = mgh) — then the oscillation movement of the candle occurs in a variation of kinetic energy (E c = mV 2 /2) — A- A
07- I. False — the electrical energy (W) generated per year is supplied by — (P o )=energy m(W)/time(∆t) — W=P o .∆t — as the time (∆t) is the same (1 year) — W=P o — according to the table the energy generated by Itaipu (93 billion kWh) is greater than the energy generated by Three Gorges (84 billion kWh), however the maximum electrical energy generation capacity, which corresponds to the installed power, is greater for Three Gorges (18,200MW) than for Itaipu (12,600MW).
II. Correct — Itaipu is more efficient because, despite having less installed capacity (12,600MW), it is capable of producing more electrical energy annually (93 billion kWh/year).
III. Correct — efficiency (η) can be the ratio between the installed power and the flooded area — Itaipu — η=12,600
R- And
08- Data — Total energy — W total = 1.6.1022 J which is the energy accumulated by the Earth in one year — energy needed to melt a mass of 1kg of ice at 0 o C, W ice = 3.2.105 J — rule of three — 1kg – 3.2.105 J — mkg – 1.6.1022 J — 3.2.105 m = 1.6.1022
— m=1.6.10 22 /3.2.10 5 — m=0.5.10 17 — m=5.0.10 16 kg — trillion=10 12 — ton=10 3 kg — trillion tons corresponds to =10 9 .10 3 =10 15 kg — m=5.10 16 kg=50.10 15 kg — m=50 trillion tons — R- B
09- Power(P o )=energy(W)/time(∆t) — W=P o .∆t — air conditioning — P oarcond = 1.5.8=12kWh —P ochuv = 3.3.(1/3)=1.1kWh.
P ofreezer =0.2.10=2kWh — P ogelad. =0.35.10=3.5kWh.
P olamp. =0.10.6=0.6kWh — total daily electrical power — P’ ot =12 + 1.1 + 2 + 3.5 + 0.6=19.2kWh — total monthly power — P ot =19.2=576kWh — monthly expenditure — G=576×0.40 — G= R$ 230.40
R- And
10- Calculation of the energy (W) consumed by the dryer with power P o = 1,000W = 1kW by the 4 people for a period of 15 min per day ∆t = 15min = 1 / 4h — P o = W / ∆t — W = P o . ∆t — W total = 4 friends x 20 days x . 1. 1 / 4 — W total = 20kWh — price of 1kWh = total spent in a month (162.50) / (monthly consumption = 260) = R $ 0.625 — spent — G = 20. 0.625 — G = R $ 12.50 — R – B
11- The statement states that at this site the height of the water tank H is four times the height of the source h — H=4h — observe in the table that for this relationship h/H=1/4, the minimum flow rate of the source for the system to function is V f =720L/h and, under these conditions, the minimum flow rate of water to be pumped is V b =120L/h — substituting these values in the efficiency expression ε=(H/h)x(V b /V f )=(4h/h)x(120/720) — ε=67% — as the efficiency of the gasoline pump is ε=36% — R- E
12- The maximum pumping flow rates of water from the tank for H=4h and H=6h are, respectively, 210 liters and 140 liters per hour — for H=5h, the flow rate will be the arithmetic mean of the values when H=4h (210L/h) and H=6h (140L/h) — V b =(210 + 140)/2=175L/h
then, the hydraulic ram will pump a maximum of 175 liters of water into the water tank — The minimum water pumping flow rates from the tank for H=4h and H=6h are, respectively, 120 liters and 80 liters per hour — for H=5h, the flow rate will be the arithmetic mean of the values when H=4h (120L/h) and H=6h (80L/h) — V b =(120 + 80)/2=100L/h
then, the hydraulic ram will pump at least 100 liters of water into the water tank — R- D
13- Note in the table that the monthly thermal energy loss due to a 10cm wall is 15kWh — the loss due to a 4cm wall is 35kWh — increase in energy loss — ∆W = 35 – 15 = 20kWh — calculation of the percentage of increase in loss in relation to the total monthly consumption — rule of three — 200kWh – 100% — 20kWh – p — 200p=20×100 — p=10%
R-C
14- Note in the graph that the shower is responsible for 25% of energy consumption — with the average monthly consumption in the residence being 300kWh, the consumption corresponding to baths due to the shower will be 25% of 300kWh, that is, 0.25×300=75kWh
Being W=P o .∆t — 75kWh=5kW — ∆t=15h — then the shower must be on for 15h per month, or 0.5h per day — as there are 4 residents, each one can use the shower per day — U=0.5h/4=60min/4 — U=7.5 min — R- C
15- Calculation of the percentage fraction of electrical energy consumption, for each appliance — F % = electrical energy (W) consumed by the appliance/total electrical energy (W total ) x 100% — as W=P o .∆t, the electrical energy consumed by a type of appliance (W appliance ) is given by W appliances = number of appliances (n) x appliance power (P o ) x appliance operating time (∆t) —
W devices =nP o . ∆t — R- E