A25.0-g sample of a metal at 98.0 °c is placed in 50.0 g of
water at 18.0 °c. if the final temperature of the metal and
water is 27.4 °c, what is the specific heat, in j/g °c, of the
metal? (3.5)

1.114 J/g°C

Mass of the metal sample = 25.0 g

Temperature change from 98°C to 27.4° C = - 70.6°C

Assuming the specific heat capacity of the metal is x, then

Heat lost = Mass × specific heat capacity × change in temp

               = 25.0 g × x × -70.6 °C

              = -1765x joules (-ve indicates that heat was lost)

              = 1765x joules

Step 2: Calculate heat gained by water

Mass of water = 50.0 g

Temperature change, from 18.0°C to 27.4°C = 9.4°C

Specific heat capacity of water = 4.184 J/g°C

Therefore,

Heat gained = 50.0 g × 4.184 J/g°C × 9.4°C

                     = 1966.48 Joules

Step 3: Calculate the specific heat of the metal

We know that heat lost is equal to heat gained.

Therefore;

Heat lost by the metal sample = heat gained by water

1765x J =1966.48 J

       x = 1.114 J/g°C

Therefore, the specific heat capacity of the metal is 1.114 J/g°C

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.