How will the amount of heat released by the electric tile change per unit of time if the spiral

| August 4, 2021 | education

How will the amount of heat released by the electric tile change per unit of time if the spiral of the tile was reduced by 2 times after repair?

According to the Joule-Lenz law, the amount of heat that will be released on the wire spiral:
Q = I ^ 2 * R * t, where I is the current strength, R is the resistance of the spiral (R = ρ * l / S, ρ is the resistivity of the spiral, l is the length of the spiral, S is the cross-sectional area of the spiral), t is time of passage of current.
Before renovation:
Q1 = I ^ 2 * R1 * t, R1 = ρ * l1 / S.
After renovation:
Q2 = I ^ 2 * R2 * t, R2 = ρ * l2 / S.
l2 = 0.5l1.
Q2 / Q1 = (I ^ 2 * (ρ * l2 / S) * t) / (I ^ 2 * (ρ * l1 / S) * t)) = l2 / l1 = 0.5l1 / l1 = 0.5 …
Answer: The amount of heat will decrease by 2 times.



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