# The electric heater operates from a network with a voltage of U = 120 V at a current of I = 5 A and in a time

**The electric heater operates from a network with a voltage of U = 120 V at a current of I = 5 A and in a time t = 20 minutes heats m = 1.5 kg of water from t1 = 16 “C to t2 = 100 ° C. Determine the efficiency of the heater.**

Given:

U = 120 Volts – voltage of the electrical network;

I = 5 Amperes – current in the electrical network;

t = 20 minutes = 1200 seconds – the time during which the heater operates;

m = 1.5 kilograms is the mass of water;

t1 = 16 ° Celsius – initial water temperature;

t2 = 100 ° Celsius – the final water temperature.

It is required to determine n (%) – the efficiency of the heater.

Determine the power of the heater:

W = U * I = 5 * 120 = 600 watts.

Let us determine the amount of heat that the heater will release during time t:

Q1 = W * t = 600 * 1200 = 720,000 Joules.

Let’s determine the amount of heat spent on heating water:

Q2 = c * m * (t2 – t1), where c = 4200 J / (kg * C) is the specific heat of water.

Q2 = 4200 * 1.5 * (100 – 16) = 4200 * 1.5 * 84 = 529200 Joules.

Then, the efficiency will be equal to:

n = Q2 / Q1 = 529200/720000 = 0.74 or 74%.

Answer: The efficiency of the heater is 74%.