Question

C.
A 35.5 g cube of aluminum initially at 48.5° C is submerged into 105.3 g of water at 15.4
What is the final temperature of both substances at thermal equilibrium?

Answer 1

Answer:

The final temperature of both substances at thermal equilibrium is 17.3°C

Explanation:

To calculate the final temperature of both substances at thermal equilibrium -:

First , we calculate the heat of A1  cube as follows -

            q= mSΔT

(where q = heat of the cube , m = mass of cube , S= specific heat of cube  {0.902j/g°C}, T = Temperature )

Putting the values given in the question ,

        $q= 32.5g\times0.902\j/g$°$C$$\times(T_f-45.8$°$C)$

            $29.315\times(T_f-45.8)J$

Now , calculate the heat of water -

         q=mSΔT

Putting values from the question ,

           $q=105.3g\times4.18j/g$°$C$$\times (T_f-15.4$°$C)$

              =$440.154\times(T_f-15.4)J$

          Now ,

Heat lost by water A1= Heat gained by water [negative sign about heat lost]

         $-29.315\times(T_f-45.8)J$ $=440.154\times(T_f-15.4)J$

            $\frac{-(T_f-45.8)J}{T_f-15.4)J} =\frac{440.154}{29.315}=15.0$

      $-T_f+45.8$°$C=15T_f-231.2$°$C$

       $(45.8+231.2)$°$C$=$16T_f$

        $16T_f=277.03$°$C$

       $T_f=\frac{277.03}{16}$ = 17.3°C

Therefore , the final temperature of both substances at thermal equilibrium is 17.3°C