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An immersion of heater must increase the temperature of water mass of 200kg from 25 degree...

Question:

An immersion of heater must increase the temperature of water mass of 200kg from 25 degree Celsius to 75 degree Celsius in 2 hours while operating voltage is 250v. what is the required resistance of the heater,

Heating of Water:

Heating of water is one of the most basic needs of human life. For this purpose, the immersion rod is utilized. It works on the principle of Joule heating. Heat energy is produced while an electric current passes through the rod containing resistance for a specific amount of time. This heat is now transferred to the water to be heated.

The formula of Joule heating is given by the following equation:

{eq}Q = I^2 R t {/eq}

where 'I' is the electric current, 'R' is the resistance and 't' is the time in seconds.

The amount required by water for heating is given by the following equation:

{eq}Q = mc \Delta T {/eq}

where 'm' is mass of water, 'c' is specific heat of water = {eq}4.18 \ J/gm, \Delta T {/eq} is the temperature difference.

Answer and Explanation:

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Given:

Mass of water, {eq}m = 200 \ Kg = 20,000 \ g {/eq}

Initial temperature, {eq}T_1 = 25^0C {/eq}

Final temperature, {eq}T_2 = 75^0C {/eq}

T...

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