Question

Use the table of specific heat capacities to solve this problem. A bar of aluminum is heated to 90.0C and then placed into 10.0L of water at 20.0 degrees * C the mixture cools to 22.4 degrees * C , what is the mass the aluminum bar? 

Answer 1

There's no table provided, but the known values of specific heat capacities are:

Water - 4.184 J/g^oC

Aluminum - 0.900 J/g^oC

The density of water is 0.998 kg/L at 20^oC, therefore the mass of water is $10.0L\times0.998kg/L=9.98kg=9980g$

Heat absorbed by water after adding an aluminum bar:

$Q=cm\Delta{T}=4.184J/g\cdot{^\circ}C\times9980g\times(22.4^\circ{C}-20.0^\circ{C})=1.00\cdot10^5J$

Heat released by the aluminum bar is equal to the heat absorbed by the water. Therefore,

$m(Al)=\frac{Q}{c\Delta{T}}=\frac{1.00\cdot10^5J}{0.900J/g\cdot^\circ{C}\times(90.0^\circ{C}-22.4^\circ{C})}=1640g=1.64kg$

Answer: 1.64 kg