Question

A 6.88 g piece of copper is heated from 20.0 degrees Celsius to 125 degrees Celsius and then submerged in 45.0 mL of water initially at 18.3 degrees Celsius. What is the final temperature of the water?

Answer 1

You haven’t specified the specific heat of copper given in your text; I shall use the value obtained off the internet which is 0.385 J/g.ºC. The specific heat of water is 4.186 J/g.ºC.

Heat lost by copper = (mass of copper)*(specific heat of copper)*(temperature change of copper) = (6.88 g)*(0.385 J/g.ºC)*(125 – 20)ºC = 278.124 J.

Let the final temperature of water be tºC. Assume the density of water to be 1.0 g/mL; therefore, the mass of 45.0 mL of water = (45.0 mL)*(1.0 g/mL) = 45.0 g.

Heat gained by water = (mass of water)*(specific heat of water)*(temperature change of water) = (45.0 g)*(4.186 J/g.ºC)*(t – 18.3)ºC = 188.37*(t – 18.3) J.

As per the principle of thermochemistry,

Heat lost by the hot substance (copper) = heat gained by the cold substance (water).

Therefore,

278.124 J = 188.37*(t – 18.3) J

====> t – 18.3 = 278.124/188.37 = 1.4765

====> t = 18.3 + 1.4765 = 19.7765 ≈ 19.78

The final temperature of water is 19.78ºC (ans).