A solid copper sphere S₁ of radius 10 cm initially at 320 K is kept in evacuated enclosure at 300 K. A second sphere S₂ of radius 32 cm initially at 370 K is in 350 K enclosure. Both have identical emissivity. Find nearest-integer ratio of their initial rates of temperature change (S₁/S₂).
Physics · Mixed Chapter
JEE Main Practice

Question

A solid copper sphere S₁ of radius 10 cm initially at 320 K is kept in evacuated enclosure at 300 K. A second sphere S₂ of radius 32 cm initially at 370 K is in 350 K enclosure. Both have identical emissivity. Find nearest-integer ratio of their initial rates of temperature change (S₁/S₂).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Radiation power of each sphere is P = εσ(4πR²)(T⁴−Tₑ⁴). Step 2 — Sphere heat capacity is mc = ρc(4πR³/3), so cooling rate |dT/dt| = 3εσ|T⁴−Tₑ⁴|/(ρcR). Step 3 — Ratio of cooling rates (sphere 1)/(sphere 2) = (R₂/R₁) × [(320⁴−300⁴)/(370⁴−350⁴)]. Step 4 — Compute 320⁴−300⁴ = 2.38576 × 10⁹ and 370⁴−350⁴ ≈ 3.73888 × 10⁹; multiply by 32/10 = 3.2 gives ≈2.04. Step 5 — Nearest integer = 2.