Question

Water at \(100^\circ\text{C}\) cools in 15 minutes to \(75^\circ\text{C}\) in a room temperature of \(25^\circ\text{C}\). Then the temperature of water after 30 minutes is

• A. \( \dfrac{400}{9}^\circ\text{C} \)
• B. \( \dfrac{526}{9}^\circ\text{C} \)
• C. \( \dfrac{335}{9}^\circ\text{C} \)
• D. \( \dfrac{175}{3}^\circ\text{C} \)

Hint:

In Newton’s law of cooling, temperature differences (not absolute temperatures) follow exponential decay.

Answer 1

The Correct Option is D

Step 1: Use Newton’s law of cooling.
\[ T-25 = (T_0-25)e^{-kt} \] Initially, \( T_0 = 100^\circ\text{C} \).

Step 2: Apply the condition at \( t=15 \) minutes.
\[ 75-25 = (100-25)e^{-15k} \] \[ 50 = 75e^{-15k} \Rightarrow e^{-15k} = \frac{2}{3} \]

Step 3: Find temperature at \( t=30 \) minutes.
\[ T-25 = 75e^{-30k} = 75\left(\frac{2}{3}\right)^2 = \frac{100}{3} \]

Step 4: Final temperature.
\[ T = 25 + \frac{100}{3} = \frac{175}{3} \]

Step 5: Conclusion.
\[ \boxed{\dfrac{175}{3}^\circ\text{C}} \]