TPP .Tugas Ke 5 Kelompok III (Persentase)ViCha Prabowo
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Transcript of TPP .Tugas Ke 5 Kelompok III (Persentase)ViCha Prabowo
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Kelompok IIITeknologi Pengolahan
Pangan
VICHA PRABOWO LAMOKI G 621 06 034
ANDI RESKI RAMADHANI G 621 06 036
WAHYU BUDI SETIAWAN G 621 06 040
RINA APRIANA G 621 06 046
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TEKSTUR DAN KADAR AIR BUAH
TOMAT
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Defenisi buah Tomat
Berasal dari Amerika tropis, ditanam sebagai tanaman buah di ladang, pekarangan
Tanaman ini tidak tahan hujan, sinar matahari terik, serta menghendaki tanah yang gembur dan subur
Buahnya buah buni, berdaging, kulitnya tipis licin mengilap
Dan dapat diolah menjadi produk yang bernilai tinggi.
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Varietas Tomat (Gycopersicum Esculentum Mill.)
Tomat Biasa (Lycopersicum commune) Tomat Apel (Lycopersicum pyriforme) Tomat Kentang (Lycopersicum
grandifolium) Tomat Keriting (Lycopersicum validum)
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Warna Buah Tomat
Hijau Kuning, Oranye Merah, Ungu (hitam) serta belang –
belang tergantung dari jenis pigmen yang dominan
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Tekstur dari buah Tomat Dapat ditentukan
dari : Dilihat Di rasa Di dengar
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Kadar air dari buah tomat
Tomat merupakan jenis tanaman yang masuk dalam kategori sayuran dan buah – buahan
Kadar air tomat yaitu 93 % Buah tomat terdiri dari 5-10% berat kering
tanpa air dan 1% kulit dan biji. Jika buah tomat dikeringkan, sekitar 50%
dari berat keringnya terdiri dari gula-gula pereduksi (terutama glukosa dan fruktosa), sisanya asam-asam organik, mineral, pigmen, vitamin dan lipid
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Kandungan Dalam Buah Tomat Terdapat vitamin C dan A Serta terdapat Likopen
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Hasil Olahan Buah Tomat produk kalengan pasta tomat sari buah (fruit
juice) saus tomat
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Cara Menghitung Kadar Air Tomat
A weet food product contains 85% water.After drying, it is found that 93% of original water has been removed. Determine (a) mass of water removed per kilogram of wet food and (b) copositro of dried food.
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Step – Step
DryerDryerFeed 85% Product…?
R = 93 %
Initial Water Content = F = 85%
Water Removed = R = 93% of original water content
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Solution
Select basis 1 kg wet food Mass of water in inlent sistem
Fwater = 85% x 1 kg = 0,85 kg
Water removed in drying = R =93 % X Fwater = 0,93 x 0,85
= 0,709 kg H2O/kg of wet food material.
Write material balance on water = Fwater – Pwater = R (a) P water
= 0,141 kg Write material balance on solid = Fsolid =
Psolid --->Psolid = 0,15kg
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…..thaNk you….