Water of volume 2 litre in a container is heated with a coil of `1kW` at `27^@C.` The lid of the container is open and energy dissipates at rate of `160J//s.` In how much time temperature will rise from `27^@C to 77^@C` Given specific heat of water is
`[4.2kJ//kg`]

Options

A. 8 min 20 sec
B. 6 sin 2 sec
C. 7 min
D. 14 min

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the mass of water
Given that the volume of water is 2 liters and the density of water is 1 kg/L, we can find the mass of the water.

\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 2 \, \text{L} \times 1 \, \text{kg/L} = 2 \, \text{kg} \]

### Step 2: Calculate the heat required to raise the temperature
We need to raise the temperature from \(27^\circ C\) to \(77^\circ C\). The change in temperature (\(\Delta T\)) is:

\[ \Delta T = 77^\circ C - 27^\circ C = 50^\circ C \]

The specific heat of water is given as \(4.2 \, \text{kJ/kg} \, \text{°C}\), which is equivalent to \(4200 \, \text{J/kg} \, \text{°C}\).

Now, we can calculate the heat required (\(Q\)) using the formula:

\[ Q = m \cdot c \cdot \Delta T \]

Substituting the values:

\[ Q = 2 \, \text{kg} \times 4200 \, \text{J/kg°C} \times 50 \, \text{°C} \]

\[ Q = 2 \times 4200 \times 50 = 420000 \, \text{J} = 4.2 \times 10^5 \, \text{J} \]

### Step 3: Calculate the net power
The power of the heating coil is \(1 \, \text{kW} = 1000 \, \text{W}\), and the energy dissipated is \(160 \, \text{J/s}\). Therefore, the net power (\(P_{\text{net}}\)) available for heating the water is:

\[ P_{\text{net}} = 1000 \, \text{W} - 160 \, \text{W} = 840 \, \text{W} \]

### Step 4: Calculate the time required to heat the water
Using the formula for power, we know that:

\[ P = \frac{Q}{t} \implies t = \frac{Q}{P_{\text{net}}} \]

Substituting the values we calculated:

\[ t = \frac{420000 \, \text{J}}{840 \, \text{W}} = 500 \, \text{s} \]

### Step 5: Convert time into minutes and seconds
To convert \(500\) seconds into minutes and seconds:

\[ 500 \, \text{s} = 8 \, \text{minutes} \, 20 \, \text{seconds} \]

### Final Answer
The time required for the temperature to rise from \(27^\circ C\) to \(77^\circ C\) is **8 minutes and 20 seconds**.

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