Question

An electric immersion heater of 1.08kW 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? ​

Answer 1

power of immersion heater, P = 1.08kW

= 1080 watt

mass of water converts into steam, m = 100g

Latent heat of vaporisation , Lv = 2256 J/g

so, heat required to produce 100g of steam, H = mLv

= 100g × 2256 J/g

= 225600 J

here,

energy consumed by immersion heater = energy gained by water to convert into steam

we know, power = energy / time

so, P × t = 225600

or, 1080 × t = 22560

or, t = 225600/1080 = 208.88 sec

hence, 208.88 sec is required to produce 100g of steam.

Answer 2

Answer:

The heat energy required to liquid water at a temperature of 100°C to steam is equal to the latent heat of vaporization of water 2260×103 J/kg. The heat energy required to liquid water at a temperature of 1000C to 100g of steam = 226 × 103 J. power of the immersion heater = 1.08×103 W, electrical energy consumed = Power × Time ⇒ Time = electrical energy consumed/ Power =226 × 103 /1.08×103 = 209s.