A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Explanation:

There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the Al be Component 1 and the H₂O be Component 2.

Data:  

For the Al:

For the water:

Calculations

(a) The relative temperature changes

(b) Final temperature

Check:

The second term has only two significant figures because ΔT₂ has only two.

It agrees to two significant figures

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.