Question

An iron rod is subjected to cycles of magnetization at the rate of 50 hz. Given the density of the rod is  and specific heat is. The rise in temperature per minute, if the area enclosed by the b – h loop corresponds to energy of  is (assume there is no radiation losses)

Answer 1

Hello dear,

● Answer-
dθ/dt = 8.12 °C/min

● Explaination-
# Given-
A = 10^-2 m^2
f = 50 Hz = 3000 bpm
d = 8×10^3 kg/m^3
c = 0.462×10^-3 J/kg°C

# Solution-
Here, heat produced per minute is given by-
dQ/dt = A × f
dQ/dt = 10^-2 × 3000
dQ/dt = 30 J

Rise in temperature per minute is now given by-
dQ/dt = dθ/dt × c × d
dθ/dt = dQ/dt / (c × d)
dθ/dt = 30 / (0.462×10^-3 × 8×10^3)
dθ/dt = 8.12 °C/min

Rise in temperature per minute is 8.12 °C/min.

Hope this is useful...