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scoundrel [369]
3 years ago
9

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 o

f 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 .)
Chemistry
1 answer:
11111nata11111 [884]3 years ago
4 0
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 





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