What amount of heat is released in 30 minutes by a wire spiral with a resistance of 20 ohms at a current of 5a.

univerkov | March 28, 2021 | education

Given:
I = 5A
R = 20 Ohm
t = 30 min = 30 * 60 = 1800s (SI system)
Find: Q -?

Decision:
According to the Joule-Lenz law:
the amount of heat released by the conductor is
Q = I ^ 2 * R * t [J]
Q = 5 ^ 2 * 20 * 1800 = 25 * 20 * 1800 = 900000 [J] = 900 [kJ]
Answer: 900 kJ

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