Question

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.

Answer 1

Answer:

Final temperature T is 26.197 degrees celcius.

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.