Question Bank • QB_APE_PROP_004 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Engineering Properties of Agriculture Produce

Subtopic: Thermal properties

Question Statement:

Strawberries with a moisture content of $90\%\text{ (wb)}$ are to be frozen. The initial temperature is $25^\circ\text{C}$ and the final temperature is $-18^\circ\text{C}$. The initial freezing point of strawberries is $-1^\circ\text{C}$. Using Siebel's equations, the specific heat above freezing is $c_{p1} = 0.837 + 3.349 X_w\text{ kJ/(kg}\cdot\text{K)}$ and below freezing is $c_{p2} = 0.837 + 1.256 X_w\text{ kJ/(kg}\cdot\text{K)}$, where $X_w$ is water mass fraction. The latent heat of freezing of water is $333.2\text{ kJ/kg}$. The total heat removed to freeze $500\text{ kg}$ of strawberries is ________ $\text{MJ}$ (round off to two decimal places).

Answer Key:

216.73

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Three-Stage Sensible and Latent Enthalpy Balance** 1. Specific heat values ($X_w = 0.90$): $c_{p1} = 0.837 + 3.349(0.90) = 0.837 + 3.0141 = 3.8511\text{ kJ/(kg}\cdot\text{K)}$ $c_{p2} = 0.837 + 1.256(0.90) = 0.837 + 1.1304 = 1.9674\text{ kJ/(kg}\cdot\text{K)}$ 2. Sensible heat removed above freezing ($25^\circ\text{C}$ to $-1^\circ\text{C}$, $\Delta T_1 = 26\text{ K}$): $Q_1 = m \cdot c_{p1} \cdot \Delta T_1 = 500 \times 3.8511 \times 26 = 50064.3\text{ kJ}$ 3. Latent heat removed at $-1^\circ\text{C}$ ($m_w = 500 \times 0.90 = 450\text{ kg}$): $Q_2 = m_w \cdot L_f = 450 \times 333.2 = 149940.0\text{ kJ}$ 4. Sensible heat removed below freezing ($-1^\circ\text{C}$ to $-18^\circ\text{C}$, $\Delta T_2 = 17\text{ K}$): $Q_3 = m \cdot c_{p2} \cdot \Delta T_2 = 500 \times 1.9674 \times 17 = 16722.9\text{ kJ}$ 5. Total heat removed: $Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 50064.3 + 149940.0 + 16722.9 = 216727.2\text{ kJ} \approx 216.73\text{ MJ}$ **Method 2: Enthalpy Per kg Unit** $q = c_{p1}(26) + X_w L_f + c_{p2}(17) = 3.8511(26) + 0.90(333.2) + 1.9674(17)$ $q = 100.1286 + 299.88 + 33.4458 = 433.454\text{ kJ/kg}$ $Q_{\text{total}} = 500 \times 433.454\text{ kJ} = 216727\text{ kJ} = 216.73\text{ MJ}$
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