Answer:
 1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block ) 
 Δs ( entropy change for cold block ) = Q / tc 
∴ Total Δs = ΔSc + ΔSh
                  = Q/tc - Q/th 
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k 
Explanation:
<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold ) 
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment 
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics ) 
Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block ) 
 Δs ( entropy change for cold block ) = Q / tc 
∴ Total Δs = ΔSc + ΔSh
                  = Q/tc - Q/th 
<u>2) Entropy change for Decomposition of mercuric oxide </u>
2HgO (s) → 2Hg(l) + O₂ (g) 
Δs = positive 
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C 
hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO = 
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k