Question

A cube's temp was increased by 1%, power (energy released per unit time) increased by 4.5%... Find increase in %volume

Answer 1

0.63%

Answer 2

Power radiated $P \propto A T^4$. Area $A \propto L^2$. Volume $V \propto L^3$.

$P/P_0 = (A/A_0) (T/T_0)^4$. $A/A_0 = (L/L_0)^2$. $V/V_0 = (L/L_0)^3$.

Given $T/T_0 = 1.01$ and $P/P_0 = 1.045$.

$1.045 = (L/L_0)^2 (1.01)^4$. $(L/L_0)^2 = 1.045 / (1.01)^4 \approx 1.045 / 1.0406 \approx 1.0042$. $L/L_0 \approx \sqrt{1.0042} \approx 1.0021$. $V/V_0 = (L/L_0)^3 \approx (1.0021)^3 \approx (1 + 0.0021)^3 \approx 1 + 3 \times 0.0021 = 1.0063$.

Percentage increase in volume $= (V/V_0 - 1) \times 100\% \approx 0.0063 \times 100\% = 0.63\%$.

The final answer is $\boxed{0.63\%}$.