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 in 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⁻¹ and that of copper = 390 J kg⁻¹°C⁻¹. The density of copper = 3.4 times the density of aluminium.

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Solution

(a) Rate of loss of heat = eAσT⁴

$\frac{{(Rate of loss of heat)}_{Al}}{{(Rate of loss of heat)}_{C}} = \frac{e_{Al} σ \cdot T^{4} \times 4 π r^{2}}{e_{C} σ T^{4} \cdot 4 π {(2 r)}^{2}} = 1 : 4 (a s e_{Al} = e_{C})$

(b) Relation between the amount of heat loss by both the spheres in a small time $∆$ t is given by

$Δ Q_{1} = 4 \times Δ Q_{2} \Rightarrow m_{1} s_{1} Δ T_{1} = 4 m_{2} s_{2} Δ T_{2} \Rightarrow \frac{Δ T_{1}}{Δ T_{2}} = \frac{4 m_{2} s_{2}}{m_{1} s_{1}} \Rightarrow \frac{Δ T_{1}}{Δ T_{2}} = \frac{4 \times ρ_{2} \times \frac{4}{3} π (2 r)^{3} \times s_{2}}{ρ_{1} \times \frac{4}{3} π r^{3} \times s_{1}} \Rightarrow \frac{Δ T_{1}}{Δ T_{2}} = \frac{4 \times 3.4 ρ_{1} \times 8 \times 390}{ρ_{1} \times 900} = 47.14 : 1$

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