in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet, the water in the kettle gained only 5.95×10^2 J of thermal energy. determine the percent efficiency of the electrical element in heating the kettle of water​

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, = 3.34 × 10³ J

The amount thermal energy gained by the kettle,  = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

Therefore, we get;

The percentage efficiency of the electrical element, η% ≈ 82.186%.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.