Lake Erie contains roughly 4.00 multiplied by 1011 m3 of water.
(a) How much energy is required to raise the temperature of that volume of water from 11.0°C to 13.0°C?
J
(b) How many years would it take to supply this amount of energy by using the 950-MW exhaust energy of an electric power plant?
I will be happy to critique your thinking.
To calculate the energy required to raise the temperature of a given volume of water, we can use the specific heat capacity formula:
Q = mcΔT
where:
Q is the energy required (in joules),
m is the mass of the water (in kilograms),
c is the specific heat capacity of water (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
(a) Energy required to raise the temperature of the water:
Given:
Volume of water = 4.00 x 10^11 m^3
Density of water = 1000 kg/m^3 (Assuming it is at 4°C)
Change in temperature (ΔT) = 13.0°C - 11.0°C = 2.0°C
To calculate the mass of the water, we can use the relationship between volume, mass, and density:
mass = volume x density
mass = (4.00 x 10^11 m^3) x (1000 kg/m^3)
Now, using the formula for energy:
Q = mcΔT
Substituting the values, we get:
Q = (4.00 x 10^11 m^3 x 1000 kg/m^3) x (2.0°C)
Simplifying the values and units,
Q = 8 x 10^14 kg x °C
So, the answer is 8 x 10^14 joules.
(b) To determine the number of years it would take to supply this energy using a 950-MW electric power plant's exhaust energy, we need to calculate the energy output of the power plant and divide it by the required energy.
Given:
Power output of the electric power plant = 950 MW (1 MW = 10^6 watts)
Time taken = t (in years)
The energy supplied by the power plant is calculated using this formula:
Energy supplied = Power x time
Converting the power output to watts:
Power = 950 MW x 10^6 watts per MW
Now, we can calculate the time taken by rearranging the formula:
t = Energy supplied / Required energy
Substituting the values, we get:
t = (Power x time) / Required energy
t = [950 MW x 10^6 watts per MW x t] / (8 x 10^14 joules)
Simplifying the values and units,
t = (950 x 10^6 W x t) / (8 x 10^14 J)
We can cancel out the units:
1 Watt (W) = 1 Joule/second (J/s)
t = (950 x 10^6 J/s x t) / (8 x 10^14 J)
t = (950/8) x (10^6/10^14) x t
t = 118.75 x 10^-8 t
Hence, the time taken in years will be equal to 118.75 times 10 to the power of -8 multiplied by t years.