A food material has a thermal conductivity $k = 0.45\text{ W/(m}\cdot\text{K)}$, density $\rho = 1050\text{ kg/m}^3$, and specific heat capacity $C_p = 3.60\text{ kJ/(kg}\cdot\text{K)}$. The thermal diffusivity ($\alpha = \frac{k}{\rho C_p}$) in $\text{m}^2\text{/s}$ is ________ $\times 10^{-7}$ (round off to 2 decimal places).
Answer Key:
1.19
Step-by-Step Solution & Explanation
1. Specific heat in SI base units: $C_p = 3.60\text{ kJ/(kg}\cdot\text{K)} = 3600\text{ J/(kg}\cdot\text{K)}$.
2. Thermal diffusivity $\alpha$:
$\alpha = \frac{k}{\rho C_p} = \frac{0.45}{1050 \times 3600} = \frac{0.45}{3.78 \times 10^6} \approx 1.1905 \times 10^{-7}\text{ m}^2\text{/s}$
Expressed as $\alpha \times 10^7 = 1.19$.