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:
(i) Find the expression for the temperature of the processor $T(t)$ given that $T(0)=85^\circ	ext{C}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Set up the differential equation. Given the rate of cooling: $$\frac{d}{dt}(T(t)) = -k(T(t) - 25)$$ Let $T$ denote $T(t)$. Separating the variables $T$ and $t$: $$\frac{dT}{T - 25} = -k dt$$ **Step 2:** Integrate both sides. $$\int \frac{dT}{T - 25} = \int -k dt$$ $$\log_e |T - 25| = -kt + C$$ **Step 3:** Use the initial condition to find $C$. Initially, at $t = 0$, the temperature $T = 85^\circ	ext{C}$. $$\log_e |85 - 25| = -k(0) + C$$ $$\log_e 60 = C$$ **Step 4:** Substitute $C$ back to find the specific expression for $T(t)$. $$\log_e (T - 25) = -kt + \log_e 60$$ $$\log_e (T - 25) - \log_e 60 = -kt$$ $$\log_e \left(\frac{T - 25}{60}ight) = -kt$$ Converting from logarithmic to exponential form: $$\frac{T - 25}{60} = e^{-kt}$$ $$T - 25 = 60e^{-kt}$$ $$T(t) = 25 + 60e^{-kt}$$ The expression for the temperature of the processor is **$T(t) = 25 + 60e^{-kt}$**.

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