Question

A 100.0g sample of tin is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 27.4 oC. Calculate the specific heat capacity of tin from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. Specific heat of water

Answer 1

Explanation:

It is known that specific heat of water is 4.184 $J/g^{o}C$ and atomic mass of tin is 118.7 g/mol. For the given situation,

                 $Q_{lost} = Q_{gained}$

Let us assume that,

               $m_{1}$ = mass of Sn

               $m_{2}$ = mass of $H_{2}O$  

Therefore, heat energy expression for heat lost and gained is as follows.

           $Q_{lost} = Q_{gained}$

      $m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})$

   $100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)$

           $7260C_{1} = 150 \times 4.184 \times 2.4$

                 $C_{1} = \frac{1506.24}{7260}$

                              = 0.207 $J/g^{o}C$

For, 118.7 g the specific heat of tin will be calculated as follows.

               $C_{1} = 0.207 J/g^{o}C \times 118.7 g$

                          = 24.5 $J/mol^{o}C$

Thus, we can conclude that specific heat of tin is 24.5 $J/mol^{o}C$.