heater coil connected to a 220 V has a resistance of 220 ohm How long will it take for this coil to heat
1 kg of water from 20 °C to 75 °C assuming that all heat is taken up by water? Specific heat capacity of water = 4200 J kg 1 °C-1
Answers
Solutions :
R=100Ω E=6v
Heat capacity of the coil 4j/k ΔT=15
0
C
Heat liberate = Increase in intarnal energy
⇒
Rt
E
2
=4J/K×15
⇒
100
6×6
×t=60⇒t=166.67sec=2.8min
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Answer:
It will take 0.3 hours for the heating coil to heat the water of 1 Kg from 20 °C to 75 °C.
Explanation:
The potential difference across the heating coil, V = 220 V
The resistance of the heating coil. R = 220 Ω
The mass of the water to be heated, m = 1 Kg
The initial temperature of the water, t₁ = 20 °C
The final temperature of the water, t₂ = 75 °C
Change in temperature, T = t₂ -t₁ = 75 - 20 = 55 °C
The specific heat capacity of water, S = 4200 J/Kg°C
Poer of the heating coil, P = V²/R
= 220² / 220
= 220 W
Heat generated = P × time
= (220 × time) J
Heat loss = mass × specific heat capacity × temperature difference
= m × S × T
= 1 × 4200 × 55
= 231000 J
It is assumed that all the heat is taken up by the water. So heat generated is equal to the heat lost.
ie; heat generated = heat loss
(220 × time) = 231000
∴ time = 231000 / 220
= 1050 s = 0.291 hr ≈ 0.3 hr