An apple contains 70% of water which has to be dried at 100°C down to moisture content of 15%. If the initial temperature of the food is 21°C, The latent heat of vaporization of water at 100°C and at standard atmospheric pressure is 2257kJkg-1. The specific heat capacity of the food is 3.8 kJkg-1°C-1 and of water is 4.186 kJkg-1°C-1. Calculate the quantity of heat energy required per unit weight of the original material, for drying under atmospheric pressure? Correct Answer 1810kJ Calculating for 1 kg food, Initial moisture = 70% 700g moisture is associated with 300 g dry matter Final moisture = 15 % 15g moisture is associated with 985 g dry matter Therefore (100 * 300)/900g = 30.6g moisture are associated with 300 g dry matter. 1kg of original matter must lose (700 -30.6) g moisture = 669.4 g = 0.669 kg moisture Heat energy required for 1kg original material = heat energy to raise temperature to 100°C+ latent heat to remove water = (100 -21) * 3.8 + 0.669 *2257 = 300.2 + 1509.9 = 1810.1kJ