Question

A dry food product has been exposed to a 30% relative-humidity environment
at 15°C for 5 h without a weight change. The moisture content has been measured and is at 7.5% (wet basis). The product is moved to a 50% relative-humidity environment, and a weight increase of 0.1 kg/kg product occurs before equilibrium is achieved.

a. Determine the water activity of the product in the first and second
environments.
b. Compute the moisture contents of the product on a dry basis in both
environments.

Answer 1

Part a

Water activity = Relative humidity / 100

In the first environment at 30% relative-humidity

Water activity = 30/100 = 0.30

In the second environment at 50% relative-humidity

Water activity = 50/100 = 0.50

Part b

In the first environment at 30% relative-humidity

Dry basis moisture content = kg water / kg solids

= (0.075) / (1 - 0.075)

= 0.0811

In the second environment at 50% relative-humidity

Kg water = (0.075) + 0.10

= 0.175

Dry basis moisture content = kg water / kg solids

= (0.175) / (1 - 0.175)

= 0.212