A 15.7-g aluminum block is warmed to 53.2 °C and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 °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?

Solution 107P

Here. we are going to calculate the final temperature of the water and Al.

When two systems are combined together, then thermal energy is transferred from hotter body to cooler one. Thus, the heat absorbed by one system is equal to the heat released by another system

Therefore,

Qsystem (1) = - Q system(2) -----(1)

Step 1:

Thermal energy transfer from metal to water.

From equation(1),

Qmetal = - Qwater

m metal x Cs metal x metal = -m water x Cs water x water -----(2)

Step 2:

Calculation of final temperature of water and Al metal.

Given,

Mass of Al metal, m metal = 15.7 g

Initial temperature of Al metal T1 = 53.2 °C

Mass of Water, m water = 32.5 g

Initial temperature of water T1 = 24.5 °C

Heat capacity,

Cs, H2O = 4.18 J/g oC ;

Cs,Al = 0.903J/g oC

From the equation (1)

m metal x Cs metal x metal = -m water x Cs water x water

15.7 g x 0.903J/g oC x metal = - 32.5 g x4.18 J/g oC x water

x metal = - 135.85 x water -------(2)