A heater coil connected to 200V has a resistance of 80 Ω. If the heater is plugged in for the time t such that 1 kg of water at 20℃ attains a temperature of 60℃. Find:
The power of heater.
The heat absorbed by the water.
The value of t in seconds.
Answers
Answer:
Explanation:
=> Here, it is given that a heater coil connected to 200V has a resistance of 80 Ω.
Voltage = 200V
Resistance = 80 ohm
=> If the heater is plugged in for the time t such that 1 kg of water at 20℃ attains a temperature of 60℃.
mass of water = 1 kg = 1000 g
initial temperature, t₁ = 20⁰C
t₂ = 60⁰C
Δt = t₂ - t₁
Δt = 60⁰ - 20⁰ = 40⁰
1) The power of heater, P = V²/ R:
= (200)²/80
=40000/80
= 500 Js⁻¹
2) The heat absorbed by the water:
Q = msΔt
=1000*4.18*40⁰
= 167200J
=167.2KJ
3) The value of t in seconds:
Q = pt
t = Q/P
= 167200/500
= 334.4 sec.
Answer:
=> Here, it is given that a heater coil connected to 200V has a resistance of 80 Ω.
Voltage = 200V
Resistance = 80 ohm
=> If the heater is plugged in for the time t such that 1 kg of water at 20℃ attains a temperature of 60℃.
mass of water = 1 kg = 1000 g
initial temperature, t₁ = 20⁰C
t₂ = 60⁰C
Δt = t₂ - t₁
Δt = 60⁰ - 20⁰ = 40⁰
1) The power of heater, P = V²/ R:
= (200)²/80
=40000/80
= 500 Js⁻¹
2) The heat absorbed by the water:
Q = msΔt
=1000*4.18*40⁰
= 167200J
=167.2KJ
3) The value of t in seconds:
Q = pt
t = Q/P
= 167200/500
= 334.4 sec.
Explanation: