Respuesta :
Explanation:
It is known that specific heat of water is 4.184 [tex]J/g^{o}C[/tex] and atomic mass of tin is 118.7 g/mol. For the given situation,
[tex]Q_{lost} = Q_{gained}[/tex]
Let us assume that,
[tex]m_{1}[/tex] = mass of Sn
[tex]m_{2}[/tex] = mass of [tex]H_{2}O[/tex]
Therefore, heat energy expression for heat lost and gained is as follows.
[tex]Q_{lost} = Q_{gained}[/tex]
[tex]m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})[/tex]
[tex]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)[/tex]
[tex]7260C_{1} = 150 \times 4.184 \times 2.4[/tex]
[tex]C_{1} = \frac{1506.24}{7260}[/tex]
= 0.207 [tex]J/g^{o}C[/tex]
For, 118.7 g the specific heat of tin will be calculated as follows.
[tex]C_{1} = 0.207 J/g^{o}C \times 118.7 g[/tex]
= 24.5 [tex]J/mol^{o}C[/tex]
Thus, we can conclude that specific heat of tin is 24.5 [tex]J/mol^{o}C[/tex].
