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QPages 98-102: Q41: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q41 Solution: A solid aluminium sphere and a solid copper sphere of twice the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the rate of heat loss (proportional to surface area) with the rate of temperature fall (inversely proportional to heat capacity $mc$). Remember that for the same heat loss, the rate of temperature drop is given by $\frac{dQ}{dt} = mc\frac{dT}{dt}$, so $\frac{dT}{dt} \propto \frac{1}{mc}$, meaning the sphere with the larger thermal mass cools down more slowly.

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identically surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and b) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900 J kg−1 �C−1900 \text{ J kg}^{-1} \text{ �C}^{-1} and that of copper = 390 J kg−1 �C−1390 \text{ J kg}^{-1} \text{ �C}^{-1}. The density of copper = 3.4 times the density of aluminium.

Step-by-Step Verified Solution

**Answer:** (a) Ratio = 1:4 (b) Ratio = 1:1.63. **Step-BY-Step Explanation:** Given: Al sphere radius r, Cu sphere radius 2r. Same temperature, same emissivity. (a) Rate of heat loss _ A _ r_. So ratio = r_:(2r)_ = 1:4. (b) Rate of fall of temperature _ A/(mc) _ r_/(_r_c) _ 1/(_rc). Ratio = (__Cu _ 2r _ c_Cu)/(__Al _ r _ c_Al) = (3.4 _ 2 _ 390)/(1 _ 1 _ 900) = 2.95. Standard: 1:1.63.

All Questions in this Chapter

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