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.

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$$
\textbf{5.1 Calculation of the net energy delivered to the water:}
$$

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  
$$
I = \frac{P}{V} \approx \frac{478.4\, \text{W}}{200\, \text{V}} \approx 2.392\, \text{A}.
$$

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\textbf{Final Answers:}
$$
$$
\text{5.1 } Q \approx 586180\, \text{J},
$$
$$
\text{5.2 } I \approx 2.39\, \text{A}.
$$
5.1 The net energy delivered to the water is approximately 586,180 joules. 5.2 The current required is approximately 2.39 amperes.

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