Respuesta :
Answer:
Therefore, the temperature of the water after 96 min (5760 s) = 279.25 K = 6.25 ℃
Explanation:
Given: Time: t₁ = 24 min = 24 × 60 s = 1440 s
Time: t₂ = 96 min = 96 × 60 s = 5760 s (∵ 1 min = 60 s)
Initial Temperature of the water: T(0) = 100℃ = 100 + 273 K = 373 K
Ambient temperature (temp of the freezer): Tₐ = 0℃ = 0 + 273 K = 273 K
Temperature of the water at time t₁: T(t₁) = 50℃ = 50 + 273 K = 323 K
(∵ 0℃ = 273 K)
Temperature of the water at time t₂: T(t₂) = ? K
According to the Newton's Law of Cooling:
[tex][T(t) - T_{a}] = [T(0) - T_{a}] e^{-kt}[/tex]
Here, k is the cooling constant
[tex]\therefore [T(t_{1}) - T_{a}] = [T(0) - T_{a}] e^{-kt_{1}}[/tex]
[tex]\Rightarrow [323\, K - 273\, K] = [373\, K - 273\, K] e^{-k(1440 sec)}[/tex]
[tex]\Rightarrow [50] = [100] e^{-k(1440 sec)}[/tex]
[tex]\Rightarrow 0.5 = e^{-k(1440 sec)}[/tex]
[tex]\Rightarrow ln(0.5)= -k \times 1440[/tex]
[tex]\Rightarrow (-0.69)= -k \times 1440[/tex]
[tex]\Rightarrow The\, cooling\, constant:\, k = 0.00048\, s^{-1}[/tex]
Now to find the temperature of water at time: t₂, we use the equation:
[tex][T(t_{2}) - T_{a}] = [T(0) - T_{a}] e^{-kt_{2}}[/tex]
[tex]\Rightarrow [T(t_{2}) - 273\, K] = [373\, K - 273\, K] e^{-(0.00048\, s^{-1})(5760\, s)}[/tex]
[tex]\Rightarrow [T(t_{2}) - 273] = [100] e^{-2.7726}[/tex]
[tex]\Rightarrow [T(t_{2}) - 273] = 100 \times 0.0625[/tex]
[tex]\Rightarrow [T(t_{2}) - 273] = 6.25 [/tex]
[tex]\Rightarrow T(t_{2}) = 273 + 6.25 = 279.25\, K[/tex]
∵ 0℃ = 273 K
∴ T(t₂) = 279.25 K = 279.25 - 273 ℃ = 6.25 ℃
Therefore, the temperature of the water after 96 min (5760 s) = 279.25 K = 6.25 ℃
