Chakapuli
 | |
| Type | Stew |
|---|---|
| Main ingredients | Lamb or veal, onions, cherry plums, dry white wine, tarragon leaves, herbs, garlic |
Chakapuli (Georgian: ჩაქაფული) is a Georgian[1][2][3][4] stew. It is considered to be one of the most popular dishes in Georgia.
Preparation
[edit]It is made from lamb chops or veal, onions, tarragon leaves, cherry plums or tkemali (cherry plum sauce), dry white wine, mixed fresh herbs (parsley, mint, dill, coriander), garlic, and salt.[5] Chakapuli can also be made with beef or mushrooms instead of lamb.[6]
Chopped lamb is boiled with white wine in a deep pan, and then the pan is placed in the oven and cooked slowly for 1.5 hours. After this process, the tkemali sauce is stirred into the lamb, and the chopped greens and garlic are added. The dish is then cooked for another 5 minutes in the oven and finally rested for 5 minutes before serving.[1][6]
See also
[edit]
References
[edit]- ^ a b Darra Goldstein, The Georgian Feast: The Vibrant Culture and Savory Food of the Republic of Georgia, p. 87
- ^ Tim Burford, Georgia, p. 74
- ^ Семенова С.В. Грузинская кухня, p. 16
- ^ Любомирова К. Постные блюда из мультиварки, p. 14
- ^ "Georgian Recipes: Chakapuli". Georgia About. Retrieved 6 March 2015.
- ^ a b "Chakapuli with lamb and wine". GeorgianJournal. Retrieved 2018-12-05.
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