Water of volume 4L in a closed container is heated with a coil of 2 kW. While water is heated, the container loses energy at a rate of 120 J/s. In how much time will the temperature of water rise from `27^(@)C` to `77^(@)C`? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Options

A. 5 min 15s
B. 3 min 7s
C. 11 min
D. 18 min

Explanation

To solve the problem step by step, we need to calculate the time required for the temperature of water to rise from 27°C to 77°C while accounting for the heat loss from the container.

### Step 1: Calculate the effective power used to heat the water.
The power of the heating coil is given as 2 kW, which is equivalent to 2000 W. However, the container loses energy at a rate of 120 J/s. Therefore, the effective power used to heat the water can be calculated as:

\[ \text{Effective Power} = \text{Power of Coil} - \text{Power Loss} \]
\[ \text{Effective Power} = 2000 \, \text{W} - 120 \, \text{W} = 1880 \, \text{W} \]

### Step 2: Calculate the mass of the water.
The volume of water is given as 4 L. We can convert this volume into cubic meters:

\[ \text{Volume} = 4 \, \text{L} = 4 \times 10^{-3} \, \text{m}^3 \]

The density of water is approximately 1000 kg/m³. Therefore, the mass of the water can be calculated as:

\[ \text{Mass} = \text{Volume} \times \text{Density} = 4 \times 10^{-3} \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 4 \, \text{kg} \]

### Step 3: Calculate the heat energy required to raise the temperature.
We need to calculate the heat energy required to raise the temperature of the water from 27°C to 77°C. The specific heat of water is given as 4.2 kJ/kg, which is equivalent to 4200 J/kg. The change in temperature (ΔT) is:

\[ \Delta T = 77°C - 27°C = 50°C \]

The heat energy (Q) required can be calculated using the formula:

\[ Q = m \cdot c \cdot \Delta T \]
\[ Q = 4 \, \text{kg} \cdot 4200 \, \text{J/kg°C} \cdot 50°C \]
\[ Q = 4 \cdot 4200 \cdot 50 = 840000 \, \text{J} \]

### Step 4: Calculate the time required to heat the water.
Now, we can find the time (t) required to provide this amount of heat energy using the effective power:

\[ t = \frac{Q}{\text{Effective Power}} = \frac{840000 \, \text{J}}{1880 \, \text{W}} \]
\[ t \approx 446.81 \, \text{s} \]

### Step 5: Convert time into minutes.
To convert seconds into minutes, we divide by 60:

\[ t \approx \frac{446.81}{60} \approx 7.45 \, \text{minutes} \]

### Final Answer:
The time required for the temperature of the water to rise from 27°C to 77°C is approximately **7.45 minutes**.

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