Answer to Question #235232 in Field Theory for Vic

Question

Calculate the required resistance of an immersion heater that will increase the
temperature of 1.50 kg of water from 10.00C to 50.00C in 10 minutes while operating at
110V.

Answer 1

$Power = \frac{V^2}{R}$

$= \frac{110^2}{R}$

$= \frac{1.21×10^4}{R}$

$Energy = power × time$

$time = 10 minutes = 600 seconds = 6×10^2 s$$Energy in Joules =\frac{ (1.21×6)×10^6}{R} =\frac{7.26 ×10^6}{R} in Joules$

Joules into water = 1.5 kg×4190 J/kg deg×(50-10) deg

= 251,400 Joules =

$=2.51×10^5$

$\frac{7.26 ×10^6 }{R }= 2.51 ×10^5$

$R = \frac{7.26 × 10^6}{2.51× 10^5}$

R = 28.92 ohms