Question:
A body cools from 80°C to 60°C in 5 minutes. Surrounding temperature is 20°C. Time to cool from 60°C to 40°C is:
A body cools from 80°C to 60°C in 5 minutes. Surrounding temperature is 20°C. Time to cool from 60°C to 40°C is:
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As a body gets closer to the surrounding temperature, its rate of cooling decreases. Therefore, it will always take more time to cover the same temperature drop at lower ranges.
Updated On: Jun 3, 2026
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The Correct Option is B
Solution and Explanation
Concept:
This problem is solved usingNewton's Law of Cooling**, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the surrounding temperature.
• Average form formula: \( \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) \)
• \( T_1, T_2 \): Initial and final temperatures of the body.
• \( T_s \): Temperature of the surroundings.
• \( t \): Time taken.
Step 1: Applying the formula for the first case.
Case 1: Body cools from 80°C to 60°C in 5 minutes. \[ \frac{80 - 60}{5} = K \left( \frac{80 + 60}{2} - 20 \right) \] \[ \frac{20}{5} = K (70 - 20) \quad \Rightarrow \quad 4 = 50K \quad \Rightarrow \quad K = \frac{4}{50} = \frac{2}{25} \quad \cdots (1) \]
Step 2: Applying the formula for the second case.
Case 2: Body cools from 60°C to 40°C in time \( t \). \[ \frac{60 - 40}{t} = K \left( \frac{60 + 40}{2} - 20 \right) \] \[ \frac{20}{t} = K (50 - 20) \quad \Rightarrow \quad \frac{20}{t} = 30K \quad \cdots (2) \]
Step 3: Substituting the value of K to find t.
Substitute \( K = \frac{2}{25} \) from equation (1) into equation (2): \[ \frac{20}{t} = 30 \times \left( \frac{2}{25} \right) \] \[ \frac{20}{t} = \frac{60}{25} \quad \Rightarrow \quad \frac{20}{t} = \frac{12}{5} \] Solving for \( t \): \[ 12t = 100 \quad \Rightarrow \quad t = \frac{100}{12} \approx 8.33 \text{ min} \]
• Average form formula: \( \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) \)
• \( T_1, T_2 \): Initial and final temperatures of the body.
• \( T_s \): Temperature of the surroundings.
• \( t \): Time taken.
Step 1: Applying the formula for the first case.
Case 1: Body cools from 80°C to 60°C in 5 minutes. \[ \frac{80 - 60}{5} = K \left( \frac{80 + 60}{2} - 20 \right) \] \[ \frac{20}{5} = K (70 - 20) \quad \Rightarrow \quad 4 = 50K \quad \Rightarrow \quad K = \frac{4}{50} = \frac{2}{25} \quad \cdots (1) \]
Step 2: Applying the formula for the second case.
Case 2: Body cools from 60°C to 40°C in time \( t \). \[ \frac{60 - 40}{t} = K \left( \frac{60 + 40}{2} - 20 \right) \] \[ \frac{20}{t} = K (50 - 20) \quad \Rightarrow \quad \frac{20}{t} = 30K \quad \cdots (2) \]
Step 3: Substituting the value of K to find t.
Substitute \( K = \frac{2}{25} \) from equation (1) into equation (2): \[ \frac{20}{t} = 30 \times \left( \frac{2}{25} \right) \] \[ \frac{20}{t} = \frac{60}{25} \quad \Rightarrow \quad \frac{20}{t} = \frac{12}{5} \] Solving for \( t \): \[ 12t = 100 \quad \Rightarrow \quad t = \frac{100}{12} \approx 8.33 \text{ min} \]
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