An immersion heater rated `1000 W, 220 V` is used to heat `0.01 m^3` of water. Assuming that the power is supplied at `220 V and 60%` of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from `15^@ C` to `40^@ C` ?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the increase in temperature (ΔT)
The water needs to be heated from 15°C to 40°C.

\[ \Delta T = 40°C - 15°C = 25°C \]

### Step 2: Calculate the mass of water (m)
We are given the volume of water as \(0.01 \, m^3\). The density of water is \(1000 \, kg/m^3\).

\[ m = \text{density} \times \text{volume} = 1000 \, kg/m^3 \times 0.01 \, m^3 = 10 \, kg \]

### Step 3: Calculate the resistance of the heater (R)
The power rating of the heater is \(1000 \, W\) and the voltage is \(220 \, V\). We can use the formula:

\[ R = \frac{V^2}{P} \]

Substituting the values:

\[ R = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4 \, \Omega \]

### Step 4: Calculate the heat required to raise the temperature of the water (Q)
Using the formula:

\[ Q = mc\Delta T \]

Where \(c\) (specific heat capacity of water) is \(4200 \, J/(kg \cdot K)\):

\[ Q = 10 \, kg \times 4200 \, J/(kg \cdot K) \times 25 \, K = 1050000 \, J \]

### Step 5: Calculate the effective power used for heating the water
Since only 60% of the power supplied is used to heat the water, we need to calculate the effective power:

\[ P_{\text{effective}} = 0.6 \times P = 0.6 \times 1000 \, W = 600 \, W \]

### Step 6: Calculate the time required to heat the water (t)
Using the formula:

\[ Q = P_{\text{effective}} \times t \]

Rearranging gives:

\[ t = \frac{Q}{P_{\text{effective}}} \]

Substituting the values:

\[ t = \frac{1050000 \, J}{600 \, W} = 1750 \, seconds \]

### Final Answer
The time required to increase the temperature of the water from 15°C to 40°C is approximately **1750 seconds** or **29.17 minutes**.

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