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. 100 s
C. 150 s
D. 200 s

Answer

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: Identify the parameters
- **Power of the heater (P)**: 836 W
- **Mass of water (m)**: 1 liter of water = 1 kg (since the density of water is 1 kg/L)
- **Specific heat capacity of water (c)**: 4180 J/(kg·°C)
- **Initial temperature (T_initial)**: 10°C
- **Final temperature (T_final)**: 40°C

### Step 2: Calculate the change in temperature (ΔT)
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 40°C - 10°C = 30°C \]

### Step 3: Calculate the energy required to heat the water (ΔQ)
Using the formula:
\[ \Delta Q = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ \Delta Q = 1 \, \text{kg} \cdot 4180 \, \text{J/(kg·°C)} \cdot 30 \, \text{°C} \]
\[ \Delta Q = 1 \cdot 4180 \cdot 30 = 125400 \, \text{J} \]

### Step 4: Relate energy to power and time
The relationship between energy, power, and time is given by:
\[ \Delta Q = P \cdot t \]
Rearranging to find time (t):
\[ t = \frac{\Delta Q}{P} \]
Substituting the values:
\[ t = \frac{125400 \, \text{J}}{836 \, \text{W}} \]
\[ t \approx 150 \, \text{s} \]

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

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