Answer:
The change in entropy of the object is [tex]ds = 10.89*10^{4} J/K[/tex]
Explanation:
From the question we are told that
The equation for the heat capacity is [tex]C = 14000\frac{J}{K} + (200 \frac{J}{K^2} )T + [3 \frac{J}{K^3} ] T^2[/tex]
The first temperature is [tex]T_1 = 290 K[/tex]
The first temperature is [tex]T_2 = 380 K[/tex]
Generally the change in entropy is mathematically represented as
[tex]ds = \int\limits^{T_2}_{T_1} {\frac{dQ}{T} } \,[/tex]
Where dQ is the change in the quantity of heat transferred with time which i mathematically represented as
[tex]Q =C dt[/tex]
Substituting this above
[tex]ds = \int\limits^{T_2}_{T_1} {\frac{C dt}{T} } \,[/tex]
Substituting for C
[tex]ds = \int\limits^{PT_2}_{T_1} {\frac{(1400 +200T +3T^2) }{T} } \, dt[/tex]
[tex]ds = 1400\ ln [T]+ 200 \frac{T^2}{T} +3 \frac{T^2}{2 T} \ \ | \left \ T_2} \atop {T_1}} \right.[/tex]
[tex]ds = 1400 ln [\frac{T_2}{T_1} ] + 200 (T_2 - T_1 ) + \frac{3}{2} (T_2^2 -T_1^2)[/tex]
Substituting values
[tex]ds = 1400 ln [\frac{380}{290} ] + 200 (380 - 290 ) + \frac{3}{2} (380^2 -290 ^2)[/tex]
[tex]ds = 10.89*10^{4} J/K[/tex]
