Using an electric heater with a resistance of R = 81 Ohm, water in a volume of V = 2.0 liters

Using an electric heater with a resistance of R = 81 Ohm, water in a volume of V = 2.0 liters is heated from a temperature of t1 = 20 ° C to boiling for a time τ = 20 minutes. Determine the efficiency of the heater if the mains voltage is U = 220 V.

We translate t = 20 minutes into seconds: t = 1200 s, the volume of water into mass: m = 2 kg, t2 = 373K, t1 = 293K.
Let’s write an expression for efficiency:
Efficiency = Q1 / Q2
Q1 is the amount of heat absorbed by water.
Q2 is the amount of heat released on the heater.
Let’s write expressions for them:
Q2 = I * U * t, where I is the current through the coil, U is the voltage, t is the operating time.
Let us express the current through the resistances:
I = U / R, substitute:
Q2 = (t * U ^ 2) / R

Q1 = c * m * (t2-t1), where c is the specific heat capacity of water c = 4200 J / (kg K), m is the mass of water, t2 and t1, respectively, the final and initial temperatures.
Let’s write an expression for efficiency:
Efficiency = Q1 / Q2 = (R * c * m * (t2-t1)) / (t * U ^ 2) = (81 * 4200 * 2 * 80) / (1200 * 220 ^ 2) = 0.93

Answer: efficiency = 0.93 = 93%