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A 26.5 −g aluminum block is warmed to 65.4 ∘C and plunged into an insulated beaker...

A 26.5 −g aluminum block is warmed to 65.4 ∘C and plunged into an insulated beaker containing 55.1 g water initially at 22.2 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.

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Answer #1

Let us denote water by symbol 1 and Aluminium by symbol 2
m1 = 55.1 g
T1 = 22.2 oC
C1 = 4.184 J/goC
m2 = 26.5 g
T2 = 65.4 oC
C2 = 0.902 J/goC
T = to be calculated

Let the final temperature be T oC
we have below equation to be used:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
26.5*0.902*(65.4-T) = 55.1*4.184*(T-22.2)
23.903*(65.4-T) = 230.5384*(T-22.2)
1563.2562 - 23.903*T = 230.5384*T - 5117.9525
T= 26.2583 oC
Answer: T= 26.3 oC

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know

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