Answer is Δs=0.3kJ/K
Since, the tank is insulated rigid tank, The process is constant volume process,
Air as perfect gas
Obtain the properties from the table. Consider th state 1. P=100kPa, T, = 15°C
s=6.830kJ/kg-K
v=0.827 m³/kg
Constant volume process,
Consider the state 2. T = 40°C, v, v = 0.827 m²/kg
5,6.890 kJ/kg-K
Calculate the change in entropy,
Δs=m(s, -s) =5(6.890-6.830)
Δs=0.3kJ/K
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