Heat energy

Energy is measured using the unit Joule (J). Temperature is measured using the unit degree Celsius °C). To increase the temperature of 1 liter of water by 1°C, 4200 J of heat energy is required.

(a) Kelly wants to increase the temperature of 2 liters of water from 25°C to 100°C.

(1) Find the amount of heat energy required.
(ii) The energy efficiency of an electric kettle A is 60%; that is, it requires 1 J of electrical energy to produce 0.6 J of heat energy. Find the amount of electrical energy kettle A requires.

(b) Another electric kettle B requires 80% of the electrical energy used by electric kettle A. Find the energy efficiency of electric kettle B.

Correct answer:

Q₁ =  630000 J
Q₂ =  1050000 J
η2 =  75 %

Step-by-step explanation:

$$\begin{gathered} c=4200\text{J/l/K} \\ V=2\text{l} \\ \Delta t=100-25=75\text{°C} \\ {Q}_1=V\cdot c\cdot \Delta t=2\cdot 4200\cdot 75=630000\text{J} \end{gathered}$$

$$\begin{gathered} \eta =60\%=\frac{60}{100}=\frac{3}{5}=0.6 \\ {Q}_1=\eta \cdot Q_2 \\ {Q}_2=Q_1/\eta =630000/0.6=1050000\text{J} \end{gathered}$$

$$\begin{gathered} r_2=80\%=\frac{80}{100}=\frac{4}{5}=0.8 \\ {Q}_3=r_2\cdot Q_2=0.8\cdot 1050000=840000\text{J} \\ {\eta }_2=100\cdot \frac{Q_1}{Q_3}=100\cdot \frac{630000}{840000}=\frac{100\cdot 630000}{840000}=\frac{63000000}{840000}=75\% \end{gathered}$$

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