Question

A 32.5 g cube of aluminum initially at 45.8C is submerged into 105.3 g of water at 15.4C. What isthe final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated)

Please show all steps, not getting the algebra used to solve this.

Answer 1

Data given : Mass of aluminum cube (ma)=32.5 grams

                  Initially temperature of aluminum (Tai) =45.8 ^oC

                  Mass of water (mw) = 105.3 grams

                  Initial temperature of water = 15.4 ^oC

Let the final temperature at equilibrium be temperature 'T'

To solve this problem, we will make use of enthalpy balance,

Here the heat given by the aluminum given by aluminum will be absorbed by water until they reach the equilibrium temperature

dH_alumium = dH_water           ..........(1)

Let mass of aluminum be 'ma'

Heat capacity of water ( Cpw) = 4.184 Joule/gram K

Heat capacity of aluminum (Cpa) = 0.9 joule/gram K

Substituting all the values in enthalpy balance equation,

ma*Cpa*(Tai-T)=mw*Cpw*(T-Twi)

32.5*0.9*(45.8-T)=105.3*4.184*(T-15.4)

32.5*0.9*45.8 - 32.5*0.9*T = 105.3*4.184*T -105.3*4.184*15.4

1339.65 - 29.25 T = 440.5752 T - 6784.858

1339.65 +6784.858 =(440.5752+29.25) T

8124.508 = 469.8252 T

T =8124.508/469.8252 = 17.2926 ^oC

Final temperature of both substances at thermal equilibrium is 17.2926 ^oC