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

Options

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

Explanation

To find 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: Calculate the energy required to heat the water
The energy required (Q) to heat water can be calculated using the formula:
\[ Q = mc\Delta T \]
where:
- \( m \) = mass of water (in kg)
- \( c \) = specific heat capacity of water (approximately \( 4180 \, \text{J/kg°C} \))
- \( \Delta T \) = change in temperature (in °C)

For 1 liter of water:
- \( m = 1 \, \text{kg} \) (since the density of water is \( 1 \, \text{kg/L} \))
- \( \Delta T = 40°C - 10°C = 30°C \)

Substituting the values:
\[ Q = 1 \, \text{kg} \times 4180 \, \text{J/kg°C} \times 30°C \]
\[ Q = 4180 \times 30 \]
\[ Q = 125400 \, \text{J} \]

### Step 2: Relate the energy to the power of the heater
The power (P) of the heater is given as 836 W. Power is defined as energy per unit time:
\[ P = \frac{Q}{t} \]
where:
- \( P \) = power (in watts)
- \( Q \) = energy (in joules)
- \( t \) = time (in seconds)

Rearranging the formula to find time:
\[ t = \frac{Q}{P} \]

### Step 3: Substitute the values to find the time
Now substituting the values we have:
\[ t = \frac{125400 \, \text{J}}{836 \, \text{W}} \]

Calculating:
\[ t = \frac{125400}{836} \]
\[ t \approx 150 \, \text{s} \]

### Final Answer
The time taken by the 836 W heater to heat one liter of water from 10°C to 40°C is approximately **150 seconds**.

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