An electric kettle with a 160 Ohm spiral was placed in a vessel containing 0.5 kg of water at 20 ° C.

An electric kettle with a 160 Ohm spiral was placed in a vessel containing 0.5 kg of water at 20 ° C. It was connected to a 220 V network after 5 minutes. the coil is turned off. What is the temperature of the water in the vessel? Disregard the loss of heat.

Given:

R = 120 Ohm – resistance of the coil of the electric boiler;

m = 0.5 kilogram is the mass of water in the vessel;

T1 = 20 ° Celsius – the initial temperature of the water in the vessel;

t = 5 minutes = 300 seconds – the operating time of the electric boiler;

U = 220 Volts – electrical circuit voltage;

c = 4200 J / (kg * C) – specific heat capacity of water.

It is required to determine T2 (degrees Celsius) – the final temperature of the water in the vessel.

Determine the power of the electric boiler:

W = I * U = U * U / R = 220 * 220/120 = 403 Watts (the result has been rounded to whole).

Then the amount of heat transferred by the boiler to the water will be equal to:

Q = W * t = 403 * 300 = 120900 Joules.

The water temperature will be equal to:

Q = c * m * (T2 – T1) = c * m * T2 – c * m * T1;

c * m * T2 = Q + c * m * T1;

T2 = (Q + c * m * T1) / (c * m) = (120900 + 4200 * 0.5 * 20) / (4200 * 0.5) =

= (120900 + 42000) / 2100 = 162900/2100 = 77.6 ° Celsius.

Answer: The final water temperature will be 77.6 ° Celsius.