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) 50s (B) 420s (C) 105s (D) 210s

Tardigrade Admin · Accepted Answer

L is the latent heat of vaporization of water; the heat required for producing 1 g of steam, L=540 cal =540 × 4.2=2268 J Energy supplied =1080 J / s Time required to boil 100 g of water =540 × 4.2 × 100 / 1080 =210 s

Q. An electric immersion heater of $1.08 kW$ is immersed in water. After the water has reached a temperature of $10 0^{\circ} C,$ how much time will be required to produce $100 g$ of steam?

50s

420s

105s

210s

$L$ is the latent heat of vaporization of water; the heat required for producing $1 g$ of steam,
$L = 540 c a l = 540 \times 4.2 = 2268 J$
Energy supplied
$= 1080 J / s$
Time required to boil $100 g$ of water
$= 540 \times 4.2 \times 100/1080$
$= 210 s$

Q. 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 100g of steam?

50s

420s

105s

210s

L is the latent heat of vaporization of water; the heat required for producing 1g of steam, L=540cal=540×4.2=2268J Energy supplied =1080J/s Time required to boil 100g of water =540×4.2×100/1080 =210s

Q. An electric immersion heater of $1.08 kW$ is immersed in water. After the water has reached a temperature of $10 0^{\circ} C,$ how much time will be required to produce $100 g$ of steam?

$L$ is the latent heat of vaporization of water; the heat required for producing $1 g$ of steam,
$L = 540 c a l = 540 \times 4.2 = 2268 J$
Energy supplied
$= 1080 J / s$
Time required to boil $100 g$ of water
$= 540 \times 4.2 \times 100/1080$
$= 210 s$