Respuesta :
Answer:
1 ) Īs ( entropy change for hot block ) = - Q / th Ā ( -ve shows heat lost to cold block )
Īs ( entropy change for cold block ) = Q / tc
ā“ Total Īs = ĪSc + ĪSh
Ā Ā Ā Ā Ā Ā Ā Ā Ā = Q/tc - Q/th
2) ĪSdecomposition = Īh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
1) To show that heat flows spontaneously from high temperature to low temperature
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 )
Īs ( entropy change for hot block ) = - Q / th Ā ( -ve shows heat lost to cold block )
Īs ( entropy change for cold block ) = Q / tc
ā“ Total Īs = ĪSc + ĪSh
Ā Ā Ā Ā Ā Ā Ā Ā Ā = Q/tc - Q/th
2) Entropy change for Decomposition of mercuric oxide
2HgO (s) ā 2Hg(l) + Oā (g)
Īs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500ā°C
hence ĪSdecomposition = Sā» Hg Ā - Ā Sā» HgO =
Īh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ĪSdecomposition = Īh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k