A student needs to determine the specific heat of a metal. She sets up a calorimeter with 100.0 mL of deionized water at room temperature, 21.0 Celsius. She heats up 50.0g of the unknown metal to 100.0 Celsius. After adding the hot metal to the water in the calorimeter the temperature equilibrated at 28.7 Celsius. What is the specific heat of the metal in J/gC?

The specific heat of the metal is 0.903 J/g°C

Explanation:

Step 1: Data given

Volume water = 100.0 mL = 0.100 L

Temperature = 21.0 °C

Mass of metal = 50.0 grams

Initial temperature of metal = 100.0 °C

Final temperature = 28.7 °C

The specific heat capacity of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = -Qgained

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal)*ΔT(metal) = -m(water) * c(water) * ΔT(water)

⇒with m(metal) = the mass of metal = 50.0 grams

⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒with ΔT(metal) = The change of temperature of the metal = T2 - T1 = 28.7 °C - 100 °C = -71.3 °C

⇒with m(water) = the mass of water = 100.0 ml * 1g/mL = 100 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the chagnge of temperature of water = 28.7 °C - 21.0 °C = 7.7 °C

50.0 * c(metal) * (-71.3) = -100 * 4.184 * 7.7 °C

-3565 * c(metal) = -3221.7

c(metal) = 0.903 J/g°C

The specific heat of the metal is 0.903 J/g°C

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.