A kettle with 2 litre water at `27^@C` is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate `160 J//s`, when its lid is open. In how much time will water heated to `77^@C` with the lid open ? (specific heat of water = `4.2 kJ//^@C-kg)`

Options

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

Explanation

To solve the problem, we need to determine the time required to heat 2 liters of water from 27°C to 77°C using a kettle with a heater that has a power of 1 kW, while accounting for heat loss to the atmosphere.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of water (V) = 2 liters
- Initial temperature (T_initial) = 27°C
- Final temperature (T_final) = 77°C
- Power of the heater (P_heater) = 1 kW = 1000 W
- Heat loss rate (P_loss) = 160 J/s
- Specific heat of water (c) = 4.2 kJ/(°C·kg) = 4200 J/(°C·kg)

2. **Convert Volume to Mass:**
Since the density of water is 1 kg/L, the mass (m) of the water can be calculated as:
\[ m = V = 2 \text{ kg} \]

3. **Calculate the Change in Temperature (ΔT):**
\[ \Delta T = T_{final} - T_{initial} = 77°C - 27°C = 50°C \]

4. **Calculate the Heat Required (Q):**
The heat required to raise the temperature of the water can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ Q = 2 \text{ kg} \cdot 4200 \text{ J/(°C·kg)} \cdot 50°C = 420000 \text{ J} = 420 \text{ kJ} \]

5. **Calculate the Effective Power Available for Heating:**
The effective power (P_effective) used for heating the water is the power of the heater minus the power lost to the atmosphere:
\[ P_{effective} = P_{heater} - P_{loss} = 1000 \text{ W} - 160 \text{ W} = 840 \text{ W} \]

6. **Calculate the Time Required (t):**
The time required to heat the water can be calculated using the formula:
\[ t = \frac{Q}{P_{effective}} \]
Substituting the values:
\[ t = \frac{420000 \text{ J}}{840 \text{ W}} = 500 \text{ seconds} \]

7. **Convert Time to Minutes:**
To convert seconds into minutes:
\[ t = \frac{500 \text{ seconds}}{60} \approx 8.33 \text{ minutes} \]
This can be expressed as 8 minutes and 20 seconds.

### Final Answer:
The time required to heat the water to 77°C with the lid open is approximately **8 minutes and 20 seconds**.