Time taken by a 836 W heater to heat one litre of water from `10^@C to 40^@C` is

Options

A. `50 s`
B. `100s`
C. `150s`
D. `200s`

Explanation

To solve the problem of determining the time taken by an 836 W heater to heat one liter of water from 10°C to 40°C, we can follow these steps:

### Step 1: Identify the given values
- Power of the heater (P) = 836 W
- Volume of water = 1 liter
- Initial temperature (T1) = 10°C
- Final temperature (T2) = 40°C
- Specific heat capacity of water (s) = 4180 J/(kg·°C)

### Step 2: Convert volume of water to mass
Since 1 liter of water has a mass of approximately 1 kg:
- Mass of water (m) = 1 kg

### Step 3: Calculate the temperature change (ΔT)
- ΔT = T2 - T1 = 40°C - 10°C = 30°C

### Step 4: Calculate the heat required to raise the temperature of the water (Q)
Using the formula for heat:
\[ Q = m \cdot s \cdot \Delta T \]
Substituting the values:
\[ Q = 1 \, \text{kg} \cdot 4180 \, \text{J/(kg·°C)} \cdot 30 \, \text{°C} \]
\[ Q = 1 \cdot 4180 \cdot 30 = 125400 \, \text{J} \]

### Step 5: Relate heat to power and time
The heat provided by the heater can also be expressed as:
\[ Q = P \cdot t \]
Where:
- Q = heat (J)
- P = power (W)
- t = time (s)

### Step 6: Set the equations equal to each other
From the previous steps, we have:
\[ P \cdot t = Q \]
Substituting the known values:
\[ 836 \, \text{W} \cdot t = 125400 \, \text{J} \]

### Step 7: Solve for time (t)
Rearranging the equation to solve for t:
\[ t = \frac{Q}{P} = \frac{125400 \, \text{J}}{836 \, \text{W}} \]
Calculating the time:
\[ t = 150 \, \text{s} \]

### Final Answer
The time taken by the heater to heat one liter of water from 10°C to 40°C is **150 seconds**.

---