A student reads on her electricity bill that the household used 350 kWh of electrical energy during the month of April. She wants to convert this value into...

Marking Scheme

• (a) The energy transferred by a 1 kW device operating for 1 hour [1]
• (b) \(E = 350 \times 1000 \times 3600\) [1]; \(E = 1.26 \times 10^9\) J [1]
• (c) Cost = \(350 \times 0.12 = \$42.00\) [1]
• (d) \(t = 350 / 2.0 = 175\) hours [1]

Explanation

(a) The kilowatt-hour (kWh) is a unit of energy (not power). It is defined as the energy transferred when a device with a power rating of 1 kilowatt (1000 W) operates for 1 hour (3600 s). Despite containing "watt" and "hour" in its name, it is a unit of energy because power multiplied by time gives energy.

(b) To convert kWh to joules, recall that:
1 kWh = 1000 W \(\times\) 3600 s = 3,600,000 J = 3.6 \(\times\) 10⁶ J
Therefore:
\(E = 350 \times 3.6 \times 10^6 = 1.26 \times 10^9\) J
This is 1.26 gigajoules. The conversion requires expressing power in watts and time in seconds because \(1\) J = \(1\) W \(\times\) 1 s.

(c) Electricity bills charge per kWh, so the cost calculation is straightforward:
Cost = energy used \(\times\) price per unit = \(350 \times 0.12 = \$42.00\)

(d) From the definition of energy in kWh: \(E = P \times t\), so \(t = E / P\).
The oven's power must be in kW: 2.0 kW.
\(t = \frac{350 \text{ kWh}}{2.0 \text{ kW}} = 175\) hours
This is about 5.8 hours per day over 30 days, which is a reasonable upper limit for oven usage consuming the entire monthly allocation.