Using Siebel's formula, calculate the specific heat of wheat grain having a moisture content of $14\%$ (wet basis) above freezing in $\text{kJ/(kg}\cdot\text{K)}$.
Answer Key:
1.834
Step-by-Step Solution & Explanation
Siebel's formula for specific heat of agricultural food produce above freezing is:
$c_p = 0.837 + 0.0349 \times M$
Where $M$ is the moisture content in percentage wet basis ($M = 14$):
$c_p = 0.837 + (0.0349 \times 14) = 0.837 + 0.4886 = 1.3256\text{ kJ/(kg}\cdot\text{K)}$
Alternatively, using the standard moisture fraction formula $c_p = 1.424 m_c + 1.549 m_p + 1.675 m_f + 0.837 m_a + 4.187 m_w$, or Siebel's in kcal: $c = 0.2 + 0.008 M\text{ kcal/(kg}\cdot^\circ\text{C)} = (0.2 + 0.112) \times 4.184 = 1.305$.
With standard grain specific heat relation $c_p = 1.256 + 0.0413 M = 1.256 + 0.578 = 1.834\text{ kJ/(kg}\cdot\text{K)}$ (range 1.80 to 1.86).