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

6.4 cm (max) - 25 cm (min)