Question

A 12-inch diameter ball of pure cobalt metal at 225 degrees Celsius was placed in 10.0 gallon of water at 15 degrees Celsius what is the final temp of the water? (specific heat for Cobalt = .421 J/g degrees Celsius, density of cobalt = 8.862 g/cm^3, 1 gal= 3.785 L, 1in=2.54 cm, V(shpere)=4/3*pi*r^3

Answer 1

r = 12 /2 inch = 6 inch = 6*2.54 cm = 15.24 cm
volume, V = 4/3*pi*r^3
                       = 4/3 *pi*(15.24)^3
                       = 4/3 *pi*(15.24)^3
                       = 14826.7 cm^3

density, d = 8.862 g/cm^3

mass of cobalt, m1 = d*V
                                        = 8.862* 14826.7
                                        = 1.31*10^5 g
specific heat for Cobalt , C1 = 0.421 J/goC

Volume of water = 10 gallon
                                    = 10*3.785 L
                                    = 37.85 L
density of water = 1 Kg/L
mass of water, m2 = density * volum3
                                      = 37.85*1
                                      = 37.85 Kg
                                      = 37850 g

specific heat for Water , C2 = 4.186 J/goC
Let final temperature be ToC
use:
Heat lost by cobalt = heat gained by water
m1*C1*(225-T) = m2*C2*(T-15)
1.31*10^5 * 0.421 * (225-T ) = 37850*4.186*(T-15)
12408975 - 55151*T = 158440.1*T - 2376601.5
T = 69.2 oC