Q110 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A bowl filled with very hot water cools from $98^\circ\text{C}$ to $86^\circ\text{C}$ in 2 minutes when the room temperature is $22^\circ\text{C}$. How long it will take to cool from $75^\circ\text{C}$ to $69^\circ\text{C}$ ?
- A$0.5$ minute
- Correct answerB$1.4$ minute
- C$1$ minute
- D$2$ minute
Solution
Using Newton's law of cooling:
$\dfrac{T_1 - T_2}{t} = K \left( \dfrac{T_1 + T_2}{2} - T_0 \right)$
For the first case:
$\dfrac{98 - 86}{2} = K \left( \dfrac{98 + 86}{2} - 22 \right)$
$6 = K (92 - 22)$
$6 = 70K \Rightarrow K = \dfrac{3}{35}$
For the second case:
$\dfrac{75 - 69}{t} = K \left( \dfrac{75 + 69}{2} - 22 \right)$
$\dfrac{6}{t} = K (72 - 22)$
$\dfrac{6}{t} = 50K$
Substituting the value of $K$:
$\dfrac{6}{t} = 50 \times \dfrac{3}{35}$
$\dfrac{6}{t} = \dfrac{30}{7}$
$t = \dfrac{42}{30} = 1.4 \text{ minutes}$
Answer: $1.4$ minute
$\dfrac{T_1 - T_2}{t} = K \left( \dfrac{T_1 + T_2}{2} - T_0 \right)$
For the first case:
$\dfrac{98 - 86}{2} = K \left( \dfrac{98 + 86}{2} - 22 \right)$
$6 = K (92 - 22)$
$6 = 70K \Rightarrow K = \dfrac{3}{35}$
For the second case:
$\dfrac{75 - 69}{t} = K \left( \dfrac{75 + 69}{2} - 22 \right)$
$\dfrac{6}{t} = K (72 - 22)$
$\dfrac{6}{t} = 50K$
Substituting the value of $K$:
$\dfrac{6}{t} = 50 \times \dfrac{3}{35}$
$\dfrac{6}{t} = \dfrac{30}{7}$
$t = \dfrac{42}{30} = 1.4 \text{ minutes}$
Answer: $1.4$ minute
Hence, the correct answer is option: 'B'.