Question

A body cools from a temperature of $60^\circ\mathrm{C}$ to $50^\circ\mathrm{C}$ in 10 minutes and $50^\circ\mathrm{C}$ to $40^\circ\mathrm{C}$ in 15 minutes. The time taken in minutes for the body to cool from $40^\circ\mathrm{C}$ to $30^\circ\mathrm{C}$ is

• A. 30
• B. 20
• C. 25
• D. 40

Hint:

When using Newton's Law of Cooling $\frac{dT}{dt} = -K(T_{avg} - T_s)$, always find the surrounding temperature $T_s$ first using the first two conditions.

Answer 1

The Correct Option is A

Step 1: Formula for Newton's Law of Cooling:
For small temperature differences or linear approximation, we use: \[ \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) \] where $T_s$ is the surrounding temperature. 
Step 2: Case 1 ($60 \to 50$ in 10 min):
\[ \frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - T_s \right) \] \[ 1 = K(55 - T_s) \quad \dots(1) \] 
Step 3: Case 2 ($50 \to 40$ in 15 min):
\[ \frac{50 - 40}{15} = K \left( \frac{50 + 40}{2} - T_s \right) \] \[ \frac{10}{15} = K(45 - T_s) \implies \frac{2}{3} = K(45 - T_s) \quad \dots(2) \] 
Step 4: Solve for $T_s$:
Divide equation (1) by (2): \[ \frac{1}{2/3} = \frac{K(55 - T_s)}{K(45 - T_s)} \] \[ \frac{3}{2} = \frac{55 - T_s}{45 - T_s} \] \[ 3(45 - T_s) = 2(55 - T_s) \] \[ 135 - 3T_s = 110 - 2T_s \] \[ T_s = 25^\circ\mathrm{C} \] 
Step 5: Find $K$:
From (1): $1 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30}$. 
Step 6: Case 3 ($40 \to 30$ in $t$ min):
\[ \frac{40 - 30}{t} = K \left( \frac{40 + 30}{2} - 25 \right) \] \[ \frac{10}{t} = \frac{1}{30} (35 - 25) \] \[ \frac{10}{t} = \frac{1}{30} (10) \] \[ \frac{10}{t} = \frac{1}{3} \implies t = 30 \text{ minutes} \]