Question

1. show that there are 3.6 × 10^6 joules in a kilowatt - hour.
2. what is the power when energy is consumed at the rate of 350 j/s?
3. how many watts are used when 7500 j of energy are consumed in 5 h?

Explanation:

Step1: Recall the conversion factor

1 kilowatt - hour (kWh) = 1000 watt - hour. Since 1 hour = 3600 s, then 1 watt - hour=1 W×3600 s. So 1 kWh = 1000 W×3600 s=3.6×10^{6} J. This shows that there are 3.6×10^{6} joules in a kilowatt - hour.

Step2: Calculate power for the second problem

Power (P) is defined as the rate of energy consumption, P = \frac{E}{t}. Given \frac{E}{t}=350 J/s, so the power P = 350 W.

Step3: Calculate power for the third problem

First, convert the time t = 5 h to seconds. t=5×3600 s = 18000 s. Given E = 7500 J. Using the power formula P=\frac{E}{t}, we substitute E = 7500 J and t = 18000 s. P=\frac{7500}{18000}W=\frac{5}{12}W≈0.42 W.

Answer:

1. Shown that 1 kWh = 3.6×10^{6} J as above.
2. 350 W
3. \frac{5}{12}W or approximately 0.42 W