A small immersion heater can be used in a car to heat a cup of water for coffee or tea. If the heater can heat 120 mL of water from 25 ∘C to 95 ∘C in 7.5 min . Assume the manufacturer's claim of 45 % efficiency.

Approximately how much current does it draw from the car's 12-V battery?

What is its resistance?

Explanation:

volume of water, V = 120 ml

mass of water = Volume x density = 120 x 1 = 120 g = 0.12 kg

Initial temperature, T1 = 25 °C

Final temperature, T2 = 95 °C

time, t = 7.5 min = 7.5 x 60 = 450 s

efficiency = 45%

Voltage, V = 12 V

specific heat of water, c = 4200 J/kg°C

Let the current is i

The energy given by the heater is used to warm the water.

Electrical energy = V i t

Thermal energy = m x c x ΔT

45 % of electrical energy = thermal energy

45 x 12 x i x 450 / 100 = 0.12 x 4200 x (95 - 25)

2430 i = 35280

i = 14.52 A

Let R be the resistance

R = V / i

R = 12 / 14.52

R = 0.83 ohm

answer; (arch); good luck

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