What is the edge length of the cube of platinum, in centimeters?

Question

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('heavy water,' specific heat capacity = 4.211 J/°C･g) at 25.5°C. The final temperature of the platinum and deuterium oxide mixture is 33.9°C. The density of platinum is 21.45 g/cm³ and the density of deuterium oxide is 1.11 g/mL. What is the edge length of the cube of platinum, in centimeters?

Answer 1

heat lost by platinum = heat gained by heavy water

q = mC∆T

For heavy water:

q = heat = ?

m = 1000 ml heavy water x 1.11 g/ml = 1110 g

C = 4.211 J/gº

∆T = change in temperature = 33.9º - 25.5º = 8.4º

For platinum:

q = mC∆T

q = q of water

m = mass = ?

C = 0.1256 J/gº

∆T = change in temperature = 200.0 - 33.9 = 166.1º

Setting heat lost by platinum = heat gained by heavy water, we have...

(m)(0.1256)(166.1) = (1110 g)(4.211 J/gº)(8.4º)

m = mass of platinum = 1882 g

Using density of platinum, we can solve for volume and then for edge of the cube.

volume of cube = 1882 g x 1cm³/21.45 g = 87.75 cm³

length of cube edge = cube root 87.75 cm³ = 4.44 cm