An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from `10^@` to `70^@`C ? Assume heat losses to the surroundings to be negligible.

Options

A. 5.5 h
B. 6.3 h
C. 7.75 h
D. 9.1 h

Explanation

To solve the problem, we need to determine how long it will take for an electric heater supplying 1.8 kW of power to heat 200 kg of water from 10°C to 70°C. We will use the formula for heat transfer and the relationship between power, energy, and time.

### Step-by-Step Solution:

1. **Identify the given values:**
- Power of the heater, \( P = 1.8 \, \text{kW} = 1.8 \times 10^3 \, \text{W} \)
- Mass of water, \( m = 200 \, \text{kg} \)
- Initial temperature, \( T_i = 10^\circ C \)
- Final temperature, \( T_f = 70^\circ C \)

2. **Calculate the change in temperature (\( \Delta T \)):**
\[ \Delta T = T_f - T_i = 70^\circ C - 10^\circ C = 60^\circ C \]

3. **Calculate the heat required (\( Q \)) using the formula:**
\[ Q = m \cdot C \cdot \Delta T \]
Here, the specific heat capacity of water (\( C \)) is approximately \( 4.18 \, \text{kJ/kg} \cdot \text{K} \) or \( 4180 \, \text{J/kg} \cdot \text{K} \).
\[ Q = 200 \, \text{kg} \cdot 4180 \, \text{J/kg} \cdot \text{K} \cdot 60 \, \text{K} \]
\[ Q = 200 \cdot 4180 \cdot 60 \]
\[ Q = 50280000 \, \text{J} = 5.028 \times 10^7 \, \text{J} \]

4. **Relate the heat to power and time:**
The relationship between heat, power, and time is given by:
\[ Q = P \cdot t \]
Rearranging for time (\( t \)):
\[ t = \frac{Q}{P} \]
Substituting the values:
\[ t = \frac{5.028 \times 10^7 \, \text{J}}{1.8 \times 10^3 \, \text{W}} \]
\[ t = \frac{5.028 \times 10^7}{1.8 \times 10^3} \]
\[ t \approx 27933.33 \, \text{s} \]

5. **Convert seconds to hours:**
\[ t \approx \frac{27933.33 \, \text{s}}{3600 \, \text{s/h}} \approx 7.75 \, \text{h} \]

### Final Answer:
It will take approximately **7.75 hours** to heat the 200 kg of water from 10°C to 70°C.