Question Bank • QB_DFE_PF_015 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

In food irradiation processing using Cobalt-60 gamma radiation, a $5.0\text{ kg}$ batch of fresh strawberries absorbs $12.5\text{ kJ}$ ($12500\text{ J}$) of ionizing radiation energy uniformly throughout its mass. The absorbed radiation dose in kiloGray ($\text{kGy}$, where $1\text{ Gy} = 1\text{ J/kg}$) is ________ (round off to one decimal place).

Answer Key:

2.5

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Direct Definition of Gray** Absorbed radiation dose ($D$): $D = \frac{\text{Absorbed Energy (J)}}{\text{Mass (kg)}} = \frac{12500\text{ J}}{5.0\text{ kg}} = 2500\text{ J/kg} = 2500\text{ Gy}$ Converting Gray to kiloGray ($1\text{ kGy} = 1000\text{ Gy}$): $D = \frac{2500\text{ Gy}}{1000\text{ Gy/kGy}} = 2.5\text{ kGy}$ **Method 2: Kilojoule Direct Division** $D = \frac{12.5\text{ kJ}}{5.0\text{ kg}} = 2.5\text{ kJ/kg} = 2.5\text{ kGy}$
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