A 10.0 g piece of hot metal at 300. °C was dropped into a 150.0 g sample of cooler room temperature water that was initially 25.0 °C. If 6.25 kJ of heat was transferred, what was the final temperature of the water?What was the specific heat of the metal?

The final temperature of water is 34.92 ⁰C

The specific heat of the metal is 2357.78 J/kg⁰C

Explanation:

Given;

mass of the hot metal, m = 10.0 g = 0.01 kg

temperature of the hot metal, = 300 °C

mass of water, = 150 g = 0.15 kg

initial temperature of the water, = 25.0 °C

heat lost by the hot metal = heat gained by water

Q = 6.25 kJ = 6250 J

let the final temperature of water = T

The final temperature of water is also the equilibrium temperature.

The specific heat of the metal is given by

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.