Question

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. The aluminum and the water are allowed to come to thermal equilibrium.Assuming that no heat is lost, what is the final temperature of the water and aluminum?

Answer 1

Answer:

The final temperature of aluminium ≈ 26.32 °C

Explanation:

Step 1: 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.

Step 2: 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

Step 3: 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