To solve this exercise we will use the concept related to heat loss which is mathematically given as
[tex]Q = mC_p \Delta T[/tex]
Where,
m = mass
[tex]C_p[/tex]= Specific Heat
[tex]\Delta T =[/tex] Change in temperature
Replacing with our values we have that
[tex]m = 0.1g[/tex]
[tex]C_p = 139J/Kg\cdot K \rightarrow[/tex] Specific heat of mercury
[tex]\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K[/tex]
Replacing
[tex]Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J[/tex]
Therefore the heat lost by mercury is 0.09J
