A current of 3 A is passed in a spiral lowered into the calorimeter for 25 minutes.
| February 22, 2021 | education
A current of 3 A is passed in a spiral lowered into the calorimeter for 25 minutes. During this time, water weighing 250 g is heated by 40 C. Calculate the resistance of the coil.
To find the resistance value of the spiral, we use the equality: I2 * R * t = Q (heat) = Cw * m * Δt, whence we express: R = Cw * m * Δt / (I2 * t).
Variables and constants: Cw – specific heat capacity of water (Cv = 4200 J / (kg * ºС)); m is the mass of water in the calorimeter (m = 250 g = 0.25 kg); Δt – temperature rise (Δt = 40 ºС); I – passed current (I = 3 A); t – the duration of the spiral (t = 25 min = 1500 s).
Calculation: R = Cw * m * Δt / (I ^ 2 * t) = 4200 * 0.25 * 40 / (3 ^ 2 * 1500) = 3.11 Ohm.
Answer: The resistance of the coil taken is 3.11 ohms.
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