Question

To 100.0 g water at 25.00 ºc in a well-insulated container is added a block of aluminum initially at 100.0 ºc. the temperature of the water once the system reaches thermal equilibrium is 28.00 ºc. what is the mass of the aluminum block? (the specific heat capacity of al is 0.900 j g–1 k–1 .)

Answer 1

when the amount of heat gained = the amount of heat loss

so, M*C*ΔTloses = M*C* ΔT gained

when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.

∴( M * C * ΔT )W = (M*C*ΔT) Al

when Mw is the mass of water = 100 g

and C the specific heat capacity of water = 4.18

and ΔT the change in temperature for water= 28-25 = 3 ° C

and ΔT the change in temperature for Al = 100-28= 72°C

and M Al is the mass of Al block

C is the specific heat capacity of the block = 0.9

so by substitution:

100 g * 4.18*3 = M Al * 0.9*72

∴ the mass of Al block is = 100 g *4.18 / 0.9*72

= 19.35 g