Question

Power and energy ​The terms power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶ J, or 250 Cal. On the other hand, power is the rate at which energy is used and is measured in watts (W; 1W=1 J/s). Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour (kWh), which is 3.6×10⁶ J. Suppose the power function of a large city over a 24-hr period is given by P(t) = E'(t) = 300 - 200 sin πt/12, where P is measured in megawatts and t=0 corresponds ​to 6:00 P.M​. (see figure). b. Burning 1 kg of coal produces about 450 kWh of energy. How many kilograms of coal are required to meet the energy needs of the city for 1 day? For 1 year?

Accepted Answer

Understand that the power function is given by $$P(t) = E'(t) = 300 - 200 \sin\left(\frac{\pi t}{12}\right)$$, where $$P(t) is in $$megawatts (MW) and $$t is in $$hours, with $$t=0 at 6$$:00 PM. To find the total energy used by the city over 1 day (24 hours), integrate the power function over the interval from $$t=0 to$$ $$t=24$$:

$$E(24) = \$$int_0$$^{24} P(t) \, dt = \$$int_0$$^{24} \left(300 - 200 \sin\left(\frac{\pi t}{12}\right)\right) dt$$

This will give the total energy in megawatt-hours (MWh). Evaluate the integral step-by-step:  
- Integrate the constant term $$300$$ over 24 hours.  
- Integrate the sine term $$-200 \sin\left(\frac{\pi t}{12}\right)$$ over 24 hours, using the substitution method if needed. After finding the total energy in megawatt-hours, convert it to kilowatt-hours (kWh) by multiplying by 1000, since 1 MW = 1000 kW. Use the given information that burning 1 kg of coal produces 450 kWh of energy. Calculate the kilograms of coal required by dividing the total energy in kWh by 450. For the yearly amount, multiply the daily coal consumption by 365.

Risposta video verificata per un problema simile:

Concetti chiave

Power as the Rate of Energy Consumption

Energy Units and Conversion

Integration to Find Total Energy from Power Function