Topic: Heat Transfer
The temperature distribution at a certain input of time in concrete slab during occurring is given by $T=3x^{2}+3x+16$ where x is in cm and T is in K. The rate of change of temperature with time is given $(\alpha=0.0003~cm^{2}/sec)$
- A. 0.009 K/sec
- B. 0.0048 K/sec
- C. $-0.0012~K/sec$
- D. -0.0018 K/sec
Answer: D. -0.0018 K/sec
Explanation: From the general heat conduction equation: $\frac{d^{2}T}{dx^{2}}=\frac{1}{\alpha}\frac{dT}{dt}$. Given $T=3x^{2}+3x+16$, differentiating twice with respect to x gives $\frac{dT}{dx}=6x+3$ and $\frac{d^{2}T}{dx^{2}}=6$. Substitute into the equation: $6=\frac{1}{0.0003}\frac{dT}{dt} \implies \frac{dT}{dt}=6 \times 0.0003 = 0.0018 K/s$. (Note: Option (d) in the paper has a negative sign, possibly due to a typo in the original text's options).