A 120 g insulated aluminum cup at 20°C is filled with 240 g of water at 100°C.?
A 120 g insulated aluminum cup at 20°C is filled with 240 g of water at 100°C.
(a) What is the total change in entropy as a result of the mixing process?
More Questions From This User See All
Answers & Comments
Verified answer
We need to find the final temperature of the system. Assuming no losses, the cup and water will end up at the final temperature T₂
Specific Heat from link
C_aluminum = 0.215
C_water = 1.00
Q_cup + Q_water = 0
M_cup C_cup ΔT_cup + M_water C_water ΔT_water = 0
120g * 0.215 * ΔT_cup + 240g * 1 * ΔT_water = 0
25.8 ΔT_cup + 240 ΔT_water = 0
25.8 (T₂ - T₁_cup) + 240 (T₂ - T₁_water) = 0
25.8 T₂ - 25.8 T₁_cup + 240 T₂ - 240 T₁_water = 0
265.8 T₂ = 25.8 T₁_cup + 240 T₁_water
T₂ = (25.8 T₁_cup + 240 T₁_water)/265.8
= (25.8 * 20°C + 240 * 100°C)/265.8
= 92.2°C
20°C = 293°K
92.2°C = 365.2°K
100°C = 373°K
ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg
= (MCΔT)_cup/T_avg + (MCΔT)_water/T_avg
= (120g * 0.215 * (365.2°K - 293°K))/((365.2°K + 293°K)/2)
+ (240g * 1 * (365.2°K - 373°K))/((365.2°K + 373°K)/2)
= 1862.76/392.1 + -187.2/369.1
= 0.588