An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100?C, how much time will be required to produce 100 g of steam?

Question

An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100?C, how much time will be required to produce 100 g of steam? (A) 210 s (B) 105 s (C) 420 s (D) 50 s

Tardigrade Admin · Accepted Answer

The heat required for producing 1 g of steam =540cal=540× 4.2J=2268J Energy given by immersion heater is =1.08 kW=1080W Now time taken in boiling 100 g of water =(2268× 100/1080)=210 sec

Q. An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100?C, how much time will be required to produce 100 g of steam?

210 s

105 s

420 s

50 s

The heat required for producing 1 g of steam $= 540 c a l = 540 \times 4.2 J = 2268 J$ Energy given by immersion heater is $= 1.08 kW = 1080 W$ Now time taken in boiling 100 g of water $= \frac{2268 \times 100}{1080} = 210 sec$

Q. An electric immersion heater of 1.08 kW is immersed in water. After the water has reached a temperature of 100?C, how much time will be required to produce 100 g of steam?

210 s

105 s

420 s

50 s

The heat required for producing 1 g of steam =540cal=540×4.2J=2268J Energy given by immersion heater is =1.08kW=1080W Now time taken in boiling 100 g of water =10802268×100​=210sec

The heat required for producing 1 g of steam $= 540 c a l = 540 \times 4.2 J = 2268 J$ Energy given by immersion heater is $= 1.08 kW = 1080 W$ Now time taken in boiling 100 g of water $= \frac{2268 \times 100}{1080} = 210 sec$