Cook at Seneca Health Care Center LLC

Job DescriptionJob DescriptionCook – Skilled Nursing Facility Pay Range $17.20-$19.67/hour (depending on experience)Location: Seneca Health Care Center – West Seneca, NYSchedule:Full-Time or Part-Time | Day and Evening Shifts Available | Includes Weekends and HolidaysCompany Overview:Seneca Health Care Center, a proud member of The McGuire Group’s five-star network of healthcare facilities, is seeking a reliable and motivated Cook to join our dietary team. If you have a passion for preparing meals with care and consistency—and want to grow in a stable healthcare environment—we want to hear from you!Position Summary:The Cook is responsible for preparing and serving meals according to menus, resident dietary needs, and food safety standards. You’ll play a key role in delivering high-quality nutrition and ensuring residents enjoy meals that are both nourishing and appealing.Key Responsibilities:Prepare hot and cold meals in accordance with standardized recipes and dietary restrictionsEnsure proper food temperatures and sanitation procedures are followedPortion and serve meals for residents and staffAssist with food prep, tray line setup, and kitchen organizationMaintain a clean and orderly kitchen and storage areasWork cooperatively with dietary aides and the Food Service director to ensure efficient meal serviceAdhere to all health codes, regulatory requirements, and infection control guidelinesQualifications:High school diploma or GED preferredPrior cooking experience in a healthcare, institutional, or commercial kitchen strongly preferredKnowledge of modified diets and safe food handlingAbility to work in a fast-paced environment and meet meal deadlinesTeam player with good communication and time management skillsWhy Join Seneca Health?Be part of a supportive and mission-driven team