One kilowatt electric heater is to be used with 220V d.c. supply.
(i) What is the current in the heater?
(ii) What is lts resistance?
(iii) what is the power dissipated in the heater?
(iv) How much heat in calories is produced per second?
(v) How many grams of water at `100^(@)C` will be converted per minute into steam at `100^(@)C`, with the heater?
Assume that the heat losses due to radiation are negligible. Latent heat of steam=540cal per gram.

Answer

Let's solve the problem step by step.

### Given:
- Power of the heater, \( P = 1 \text{ kW} = 1000 \text{ W} \)
- Voltage supply, \( V = 220 \text{ V} \)

### (i) What is the current in the heater?

To find the current, we can use the formula for power:
\[ P = V \times I \]
Rearranging this formula to solve for current \( I \):
\[ I = \frac{P}{V} \]
Substituting the known values:
\[ I = \frac{1000 \text{ W}}{220 \text{ V}} \approx 4.545 \text{ A} \]
Thus, the current in the heater is approximately \( 4.55 \text{ A} \).

### (ii) What is its resistance?

We can use the power formula again, but this time in terms of resistance:
\[ P = \frac{V^2}{R} \]
Rearranging to find resistance \( R \):
\[ R = \frac{V^2}{P} \]
Substituting the known values:
\[ R = \frac{(220 \text{ V})^2}{1000 \text{ W}} = \frac{48400}{1000} = 48.4 \, \Omega \]
Thus, the resistance of the heater is \( 48.4 \, \Omega \).

### (iii) What is the power dissipated in the heater?

The power dissipated in the heater is equal to the power rating of the heater, which is given as:
\[ P = 1000 \text{ W} \]
Thus, the power dissipated in the heater is \( 1000 \text{ W} \).

### (iv) How much heat in calories is produced per second?

To convert the power from watts to calories per second, we use the conversion factor \( 1 \text{ W} = 0.239 \text{ cal/s} \):
\[ \text{Heat per second} = P \times 0.239 \text{ cal/s} \]
Substituting the values:
\[ \text{Heat per second} = 1000 \text{ W} \times 0.239 \text{ cal/s} \approx 239 \text{ cal/s} \]
Thus, the heat produced per second is approximately \( 239 \text{ cal/s} \).

### (v) How many grams of water at \( 100^\circ C \) will be converted per minute into steam at \( 100^\circ C \)?

First, we calculate the total heat produced in one minute:
\[ \text{Total heat in one minute} = 239 \text{ cal/s} \times 60 \text{ s} = 14340 \text{ cal} \]
Next, we use the latent heat of steam, which is \( L = 540 \text{ cal/g} \):
Using the formula \( H = M \times L \), we can find the mass \( M \):
\[ M = \frac{H}{L} = \frac{14340 \text{ cal}}{540 \text{ cal/g}} \approx 26.58 \text{ g} \]
Thus, approximately \( 26.58 \text{ g} \) of water will be converted into steam per minute.

### Summary of Answers:
1. Current in the heater: \( 4.55 \text{ A} \)
2. Resistance of the heater: \( 48.4 \, \Omega \)
3. Power dissipated in the heater: \( 1000 \text{ W} \)
4. Heat produced per second: \( 239 \text{ cal/s} \)
5. Mass of water converted to steam per minute: \( 26.58 \text{ g} \)