An electric water heater operating at a voltage of 240\,\text{V} has a heating element of … — JAMB UTME

An electric water heater operating at a voltage of $240\,\text{V}$ has a heating element of resistance $48\,\Omega$ . It is used to heat $1.5\,\text{kg}$ of water from $20^\circ\text{C}$ to $100^\circ\text{C}$ . If the thermal efficiency of the heating process is $80\%$ , calculate the total electrical energy consumed by the heater in kilojoules ( $\text{kJ}$ ). [Take specific heat capacity of water = $4200\,\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ ]

Answer: 630 kJ

Answer

The total electrical energy consumed by the heater is

630\,\text{kJ}

The thermal energy required to raise the temperature of

1.5\,\text{kg}

of water by

80^\circ\text{C}

Q = 1.5 \times 4200 \times 80 = 504,000\,\text{J} = 504\,\text{kJ}

. Taking into account the

80\%

thermal efficiency, the total electrical energy consumed is

E_{\text{elec}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

Step-by-Step Solution

Calculate the useful heat energy needed to heat the water.

Q = m c (T_2 - T_1) = 1.5 \times 4200 \times (100 - 20) = 504,000\,\text{J} = 504\,\text{kJ}

The thermal energy transferred to the water depends on its mass, specific heat capacity, and temperature increase.

Account for the efficiency of the heating element to find total electrical energy input.

E_{\text{elec}} = \frac{Q}{\text{Efficiency}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

Since only

80\%

of the electrical energy is converted into useful heat energy for the water, the input electrical energy must be greater than the output heat energy.

Key Concept

Conversion of electrical energy to thermal energy and application of thermal efficiency.

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