Question

Boiling water at 100℃ is placed in a freezer at 0℃. The temperature of the water is 50℃ after 24 minutes. Approximate the temperature of the water after 96 minutes.

Answer 1

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:

$[T(t) - T_{a}] = [T(0) - T_{a}] e^{-kt}$

Here, k is the cooling constant

$\therefore [T(t_{1}) - T_{a}] = [T(0) - T_{a}] e^{-kt_{1}}$

$\Rightarrow [323\, K - 273\, K] = [373\, K - 273\, K] e^{-k(1440 sec)}$

$\Rightarrow [50] = [100] e^{-k(1440 sec)}$

$\Rightarrow 0.5 = e^{-k(1440 sec)}$

$\Rightarrow ln(0.5)= -k \times 1440$

$\Rightarrow (-0.69)= -k \times 1440$

$\Rightarrow The\, cooling\, constant:\, k = 0.00048\, s^{-1}$

Now to find the temperature of water at time: t₂, we use the equation:

$[T(t_{2}) - T_{a}] = [T(0) - T_{a}] e^{-kt_{2}}$

$\Rightarrow [T(t_{2}) - 273\, K] = [373\, K - 273\, K] e^{-(0.00048\, s^{-1})(5760\, s)}$

$\Rightarrow [T(t_{2}) - 273] = [100] e^{-2.7726}$

$\Rightarrow [T(t_{2}) - 273] = 100 \times 0.0625$

$\Rightarrow [T(t_{2}) - 273] = 6.25$

$\Rightarrow T(t_{2}) = 273 + 6.25 = 279.25\, K$

∵ 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 ℃