CBSE 2025, Set 3

Question

**Case Study - 3**
During a heavy gaming session, the temperature of a student's laptop processor increases significantly. After the session, the processor begins to cool down, and the rate of cooling is proportional to the difference between the processor's temperature and the room temperature ($25^\circ	ext{C}$). Initially the processor's temperature is $85^\circ	ext{C}$. The rate of cooling is defined by the equation $\frac{d}{dt}(T(t))=-k(T(t)-25)$,
where $T(t)$ represents the temperature of the processor at time $t$ (in minutes) and $k$ is a constant.

Based on the above information, answer the following question:
(ii) How much time will it take for the processor's temperature to reach $40^\circ	ext{C}$? Given that $k=0.03$, $\log_e 4 = 1.3863$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Use the temperature expression derived in Part (i). We have $T(t) = 25 + 60e^{-kt}$. **Step 2:** Substitute the given values to find $t$. We need to find the time $t$ when $T(t) = 40^\circ	ext{C}$. We are also given $k = 0.03$. $$40 = 25 + 60e^{-0.03t}$$ $$40 - 25 = 60e^{-0.03t}$$ $$15 = 60e^{-0.03t}$$ Divide by 60: $$e^{-0.03t} = \frac{15}{60}$$ $$e^{-0.03t} = \frac{1}{4}$$ **Step 3:** Solve for $t$ using logarithms. Take the natural logarithm ($\log_e$) on both sides: $$\log_e(e^{-0.03t}) = \log_e\left(\frac{1}{4}ight)$$ $$-0.03t = \log_e 1 - \log_e 4$$ Since $\log_e 1 = 0$: $$-0.03t = -\log_e 4$$ $$0.03t = \log_e 4$$ We are given $\log_e 4 = 1.3863$: $$0.03t = 1.3863$$ $$t = \frac{1.3863}{0.03}$$ $$t = \frac{138.63}{3} = 46.21$$ It will take **$46.21$ minutes** for the processor's temperature to reach $40^\circ	ext{C}$.

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