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A 27.5 −g aluminum block is warmed to 65.6 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.0 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.

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Answer:

Final temperature T is 26.197 degrees celcius.

Step-by-step explanation:

As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.

specific heat capacity of aluminium = 0.9 J/g per degree celcius

specific heat capacity of water = 4.186 J/g per degree celcius

Let the final common temperature attained be "T".

Heat lost by aluminium =
(m)(c)(65.6-T)

Heat gained by water =
(m)(c)(T-22)

by law of conservation of energy,

Heat lost by hot body = Heat gained by cold body


(m)(c)(65.6-T) =
(m)(c)(T-22)


(27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)

calculating the value of T comes out as 26.197 degrees celcius.

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