Question

A glass of hot water cools from 90°C to 70°C in 3 minutes when the temperature of the surroundings is 20°C. What is the time taken by the glass of hot water to cool from 60°C to 40°C if the surrounding temperature remains the same at 20°C?

• A. 15 minutes
• B. 6 minutes
• C. 12 minutes
• D. 8 minutes

Hint:

Use Newton's Law of Cooling to calculate the time taken for temperature changes in cooling problems. Remember that the time depends on the logarithmic difference between initial and final temperatures.

Answer 1

The Correct Option is C

We will use Newton's Law of Cooling, which states that the rate of change of temperature of the object is directly proportional to the difference between the temperature of the object and the surrounding temperature. Mathematically: \[ \frac{dT}{dt} = -k (T - T_{\text{env}}) \] Where:
- \( T \) is the temperature of the object, 
- \( T_{\text{env}} \) is the temperature of the surroundings, 
- \( k \) is the cooling constant. The time taken to cool from \( T_1 \) to \( T_2 \) is given by: \[ t = \frac{1}{k} \ln \left( \frac{T_1 - T_{\text{env}}}{T_2 - T_{\text{env}}} \right) \] 
Step 1: Calculate the cooling constant \( k \) using the first part of the cooling process:
We are given that the temperature drops from 90°C to 70°C in 3 minutes, and the surrounding temperature is 20°C. Substituting into the equation: \[ 3 = \frac{1}{k} \ln \left( \frac{90 - 20}{70 - 20} \right) \] Simplifying: \[ 3 = \frac{1}{k} \ln \left( \frac{70}{50} \right) \] \[ 3 = \frac{1}{k} \ln(1.4) \] \[ k = \frac{\ln(1.4)}{3} \] 
Step 2: Use the value of \( k \) to calculate the time taken for the temperature to cool from 60°C to 40°C:
Now, substitute into the equation for cooling time: \[ t = \frac{1}{k} \ln \left( \frac{60 - 20}{40 - 20} \right) \] \[ t = \frac{1}{k} \ln \left( \frac{40}{20} \right) = \frac{1}{k} \ln(2) \] Substitute the value of \( k \): \[ t = \frac{\ln(2)}{\ln(1.4)} \times 3 \approx 12 \text{ minutes}. \] Thus, the time taken to cool from 60°C to 40°C is 12 minutes.