Q160 / 180 · COMEDK 2024 (Morning Shift) · Physics
A glass of hot water cools from $90^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 3 minutes when the temperature of surroundings is $20^{\circ} \mathrm{C}$. What is the time taken by the glass of hot water to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ if the surrounding temperature remains the same at $20^{\circ} \mathrm{C}$ ?
- A15 minutes
- Correct answerB6 minutes
- C12 minutes
- D10 minutes
Solution
According to Newton's law of cooling, the rate of cooling is given by $\dfrac{dT}{dt} = -k(T - T_s)$, where $T$ is the temperature of the body, $T_s$ is the temperature of the surroundings, and $k$ is a constant.
For the first interval, $T_1 = 90^{\circ} \text{C}$, $T_2 = 70^{\circ} \text{C}$, $T_s = 20^{\circ} \text{C}$, and $t = 3 \text{ minutes}$. Using the average temperature approximation $\dfrac{T_1 - T_2}{t} = k \left( \dfrac{T_1 + T_2}{2} - T_s \right)$, we get $\dfrac{90 - 70}{3} = k \left( \dfrac{90 + 70}{2} - 20 \right)$.
$\dfrac{20}{3} = k(80 - 20) = 60k$, which implies $k = \dfrac{20}{3 \times 60} = \dfrac{1}{9}$.
For the second interval, $T_1 = 60^{\circ} \text{C}$, $T_2 = 40^{\circ} \text{C}$, $T_s = 20^{\circ} \text{C}$, and we need to find $t'$. Using the same formula, $\dfrac{60 - 40}{t'} = k \left( \dfrac{60 + 40}{2} - 20 \right)$.
$\dfrac{20}{t'} = \dfrac{1}{9} (50 - 20) = \dfrac{30}{9} = \dfrac{10}{3}$.
Solving for $t'$, $t' = \dfrac{20 \times 3}{10} = 6 \text{ minutes}$.
Answer: 6 minutes
For the first interval, $T_1 = 90^{\circ} \text{C}$, $T_2 = 70^{\circ} \text{C}$, $T_s = 20^{\circ} \text{C}$, and $t = 3 \text{ minutes}$. Using the average temperature approximation $\dfrac{T_1 - T_2}{t} = k \left( \dfrac{T_1 + T_2}{2} - T_s \right)$, we get $\dfrac{90 - 70}{3} = k \left( \dfrac{90 + 70}{2} - 20 \right)$.
$\dfrac{20}{3} = k(80 - 20) = 60k$, which implies $k = \dfrac{20}{3 \times 60} = \dfrac{1}{9}$.
For the second interval, $T_1 = 60^{\circ} \text{C}$, $T_2 = 40^{\circ} \text{C}$, $T_s = 20^{\circ} \text{C}$, and we need to find $t'$. Using the same formula, $\dfrac{60 - 40}{t'} = k \left( \dfrac{60 + 40}{2} - 20 \right)$.
$\dfrac{20}{t'} = \dfrac{1}{9} (50 - 20) = \dfrac{30}{9} = \dfrac{10}{3}$.
Solving for $t'$, $t' = \dfrac{20 \times 3}{10} = 6 \text{ minutes}$.
Answer: 6 minutes
Hence, the correct answer is option: 'B'.