Answer to Question #327335 in Electric Circuits for Deib

Question

A coil connected to a 220 V line is used to heat 500.0 g of water at 30°C. It is found out that after 5 minutes, the temperature of the water has risen to 80°C. What is the resistance of the coil? 

Answer 1

The electrical energy received from the coil equals:

$E=Pt=\frac{U^2}{R}t$ ,

where P - electrical power;

U - voltage;

R - resistance;

t - time.

So,

$R=\frac{U^2t}{E}.$

Energy, required for heating of the water with mass m from the temperature $T_1$ to the temperature $T_2$ equals:

$E=cm(T_2-T_1),$

where c - specific heat capacity of the water.

So,

$R=\frac{U^2t}{cm(T_2-T_1)}=\frac{220^2\cdot 300}{4184\cdot 0.5\cdot (80-30)}=138.8\space Ohm.$