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QPages 98-102: Q55: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q55 Solution: A hot body placed in a surrounding of temperature obeys…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently drop the surrounding temperature term and solve $\frac{d\theta}{dt}=-K\theta$ as if the body cools to zero, writing $\theta=\theta_1e^{-Kt}$. The correct equation is $\frac{d\theta}{dt}=-K(\theta-\theta_0)$, giving $\theta(t)=\theta_0+(\theta_1-\theta_0)e^{-Kt}$; forgetting the $\theta_0$ shift leads to an incorrect $Q_{\max}$ and wrong time for 90% heat loss.

A hot body placed in a surrounding of temperature θ0\theta_0 obeys Newton's law of cooling dθdt=−Kθ−θ0)\frac{d\theta}{dt} = -K\theta - \theta_0). Its temperature at t = 0 is θ1\theta_1. The specific heat capacity of the body is s and its mass is m. Find a) the maximum heat that the body can lose and b) the time starting from t = 0 in which it will lose 90% of this maximum heat.

Step-by-Step Verified Solution

**Answer:** (a) Q_max = ms(_____) (b) t = (ms/k) ln 10. **Step-BY-Step Explanation:** (a) Maximum heat loss occurs when body reaches surrounding temperature: Q = ms(_____). (b) For 90% heat loss, ____ = 0.1(_____). Using Newton's law: t = (1/k) ln 10.

All Questions in this Chapter

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