# Problem: A 26.0 -g aluminum block is warmed to 65.5 oC and plunged into an insulated beaker containing 55.3 g water initially at 22.2 oC. 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?

###### FREE Expert Solution

We’re being asked to determine the final temperature of aluminum and water at thermal equilibrium.

Recall that heat can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{∆}}{\mathbf{T}}}$

q = heat, J

• +q → absorbs heat
•
–q → loses heat

m = mass (g)
c = specific heat capacity = J/(g·°C)
ΔT = Tf – Ti = (°C)

Based on the given system:

• The 26.0 g of aluminum at an initial temperature of 65.5 °C was submerged into water

• The water weighs 55.3 g at  22.2 °C

• Assuming that no heat is lost to the surroundings, in order to reach thermal equilibrium, aluminum will lose heat (since it has a higher temperature) and water will absorb this heat.

▪ the heat lost by aluminum is absorbed by water

▪ heat is lost by aluminum- q

▪ heat is absorbed by water+ q

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###### Problem Details

A 26.0 -g aluminum block is warmed to 65.5 oC and plunged into an insulated beaker containing 55.3 g water initially at 22.2 oC. 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?