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 1 kW heater while accounting for heat loss to the surroundings.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Volume of water, \( V = 2 \) liters = \( 2000 \) grams (since 1 liter of water has a mass of approximately 1000 grams).
- Initial temperature, \( T_i = 27^\circ C \).
- Final temperature, \( T_f = 77^\circ C \).
- Power of the heater, \( P = 1 \) kW = \( 1000 \) J/s.
- Heat loss to surroundings, \( P_{loss} = 160 \) J/s.
2. **Calculate the Net Power:**
The net power available for heating the water can be calculated by subtracting the heat loss from the power of the heater.
\[
P_{net} = P - P_{loss} = 1000 \, \text{J/s} - 160 \, \text{J/s} = 840 \, \text{J/s}
\]
3. **Calculate the Temperature Change:**
The change in temperature, \( \Delta T \), is given by:
\[
\Delta T = T_f - T_i = 77^\circ C - 27^\circ C = 50^\circ C
\]
4. **Use the Formula for Heat Transfer:**
The heat required to raise the temperature of the water can be calculated using the formula:
\[
Q = mc\Delta T
\]
where:
- \( m = 2000 \) grams (mass of water),
- \( c = 4.2 \) J/g°C (specific heat capacity of water),
- \( \Delta T = 50 \)°C.
Plugging in the values:
\[
Q = 2000 \, \text{g} \times 4.2 \, \text{J/g°C} \times 50 \, \text{°C} = 420000 \, \text{J}
\]
5. **Calculate the Time Required:**
The time \( t \) required to provide this amount of heat with the net power can be calculated using:
\[
t = \frac{Q}{P_{net}} = \frac{420000 \, \text{J}}{840 \, \text{J/s}} = 500 \, \text{s}
\]
6. **Convert Time to Minutes:**
To convert seconds to minutes:
\[
t_{minutes} = \frac{500 \, \text{s}}{60} \approx 8.33 \, \text{minutes}
\]
This can be expressed as 8 minutes and 20 seconds.
### Final Answer:
The time required for the temperature to reach 77°C is approximately **8 minutes and 20 seconds**.
