How much Kelvin will the resistor heat up in 5 minutes (the specific resistance of which is 50 * 10 -8 degrees Ohm

| June 17, 2021 | education

How much Kelvin will the resistor heat up in 5 minutes (the specific resistance of which is 50 * 10 -8 degrees Ohm * m, the density is 5000 kg / m3), if the current density in it is 5 kA / m2?

Given: t (time) = 5 min (300 s); ρ (specific electrical resistance of the resistor) = 50 * 10 ^ -8 Ohm * m; ρr (resistor density) = 5000 kg / m3; j (current density) = 5 kA / m2 = 5000 A / m2; C (specific heat capacity of the resistor) = 500 J / (K * kg).

The change in temperature is determined from the equality: C * m * Δt = Q = I ^ 2 * R * t = j ^ 2 * S ^ 2 * ρ * l / S * t = j ^ 2 * S * ρ * l * t = j ^ 2 * ρ * V * t = j ^ 2 * ρ * m / ρр * t; Δt = j ^ 2 * ρ * m * t / (ρр * C * m) = j ^ 2 * ρ * t / (ρр * C).

Calculation: Δt = 50002 * 50 * 10 ^ -8 * 300 / (5000 * 500) = 0.0015 K.



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