QUESTION 5 Within 30 minutes, 2 liters of water must be heated from \( 10^{\circ} \mathrm{C} \) to \( 80^{\circ} \mathrm{C} \), The specific heat capacity of water is \( 4,187 \mathrm{~kJ} / \mathrm{kg} \). K . The efficiency of the heating system in \( 68 \% \). 5.1 What is the net amount of energy delivered to the water? 5.2 The voltage supply is 200 V . Determine the value of the current.

\[ \textbf{5.1 Calculation of the net energy delivered to the water:} \] 1. \(\textbf{Mass of water:}\) Since the density of water is approximately \(1\, \text{kg/L}\), 2 liters of water have a mass of \[ m = 2\, \text{kg}. \] 2. \(\textbf{Temperature change:}\) The water is heated from \(10^{\circ}\text{C}\) to \(80^{\circ}\text{C}\), so the change in temperature is \[ \Delta T = 80^{\circ}\text{C} - 10^{\circ}\text{C} = 70\, \text{K}. \] 3. \(\textbf{Specific heat capacity:}\) The specific heat capacity of water is given as \[ c = 4187\, \frac{\text{J}}{\text{kg}\cdot\text{K}}. \] 4. \(\textbf{Net energy delivered:}\) Using the formula \[ Q = m \cdot c \cdot \Delta T, \] we have \[ Q = 2\, \text{kg} \times 4187\, \frac{\text{J}}{\text{kg}\cdot\text{K}} \times 70\, \text{K} \approx 586180\, \text{J}. \] \[ \textbf{5.2 Determination of the current:} \] 1. \(\textbf{Accounting for system efficiency:}\) The heating system operates at an efficiency of \(68\%\) (\(\eta = 0.68\)). This means that in order to deliver \(Q\) joules of energy to the water, the total energy supplied electrically must be \[ E_{\text{supplied}} = \frac{Q}{\eta} = \frac{586180\, \text{J}}{0.68} \approx 861147\, \text{J}. \] 2. \(\textbf{Time available for heating:}\) The time given is 30 minutes, which in seconds is \[ t = 30\, \text{min} \times 60\, \frac{\text{s}}{\text{min}} = 1800\, \text{s}. \] 3. \(\textbf{Power required:}\) The power \(P\) needed to supply this energy in the given time is \[ P = \frac{E_{\text{supplied}}}{t} \approx \frac{861147\, \text{J}}{1800\, \text{s}} \approx 478.4\, \text{W}. \] 4. \(\textbf{Current calculation from the voltage supply:}\) Given the supply voltage \(V = 200\, \text{V}\), the current \(I\) drawn is given by \[ I = \frac{P}{V} \approx \frac{478.4\, \text{W}}{200\, \text{V}} \approx 2.392\, \text{A}. \] \[ \textbf{Final Answers:} \] \[ \text{5.1 } Q \approx 586180\, \text{J}, \] \[ \text{5.2 } I \approx 2.39\, \text{A}. \]