Question

A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100^0C to 99^0C in 100s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from 100^0C to 99^0C will be approximately

• A. 25 s
• B. 50 s
• C. 200 s
• D. 400 s

Answer 1

Answer: (C)

200 s

Answer 2

$\frac{dT}{dt} = \frac{eA\sigma}{mc}\left( T^{4} - T_{0}^{4} \right) = \frac{e(6a^{2})\sigma}{(a^{3} \times \rho)c}(T^{4} - T_{0}^{4})$ ⇒ For the same fall in temperature, time $dt \propto a$

$\frac{dt_{2}}{dt_{1}} = \frac{a_{2}}{a_{1}} = \frac{2cm}{1cm}$ ⇒ dt₂ = 2 × dt_­1 = 2 × 100 sec = 200 sec

[As A = 6a² and m = V × ρ = a³ × ρ]