Answer:
The final temperature of aluminium ≈ 26.32 °C
Explanation:
<u>Step 1:</u> explain the problem
A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.
<u>Step 2:</u> Data given
We will use the formule : Q = mcΔT
with Q = heat transfer ( J)
with m = mass of the substance (g)
with c = specific heat ( J/g °C)
with ΔT = change in temperature ( in °C or K)
mass of aluminium = 27.5g
mass of water = 55.5g
specific heat of aluminium = 0.900J/g °C
specific heat of water = 4.186 J/g °C
initial temperature of aluminium T1= 65.9 °C
initial temperature of water T1 = 22.1 °C
final temperature of water and aluminium = TO BE DETERMINED
<u>Step 3:</u> Calculate the initial temperature
To find the final temperature, we have to use the following formule:
-(Mass of aluminium) * (caluminium)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)
-27.5g (0.900)(T2 - 65.9) = 55.5g (4.184j/g °C) (T2- 22.1)
-24.75*(T2-65.9) = 232.212 *(T2-22.1)
-24.75T2 + 1631.025 = 232.212T2 -5131,8852
-256.962 T2 = -6762.9102
T2 = 26.32 °C
The final temperature of aluminium ≈ 26.32 °C
