Question

A 25.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.2 g water initially at 22.1 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.

Answer 1

Answer:

Hi, this question is incomplete. The complete question is "A 25.5-g aluminum block is warmed to $65.9^{0}\textrm{C}$ and plunged into an insulated beaker containing 55.2 g water initially at 22.1^{0}\textrm{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?"

Final temperature is 26.1^{0}\textrm{C}

Explanation:

Let's assume final temperature is T where $65.9^{0}\textrm{C}> T > 22.1^{0}\textrm{C}$

As no heat is lost therefore in equilibrium-

amount of heat released by Al block = amount of heat consumed by water

or, $m_{Al}\times C_{Al}\times \Delta T_{Al}=m_{water}\times C_{water}\times \Delta T_{water}$

Here m represents mass, C represents specific heat and $\Delta T$ represents change in temperature

So, $(25.5g)\times (0.90J/g.^{0}\textrm{C})\times (65.9^{0}\textrm{C}-T)=(55.2g)\times (4.184J/g.^{0}\textrm{C})\times (T-22.1^{0}\textrm{C})$

or, $T=26.1^{0}\textrm{C}$

So, the final temperature is $26.1^{0}\textrm{C}$