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ch14 concept review heat and temperature answers required to raise the temperature of 200 g of water from 20°C to 100°C? Given: m = 200 g c = 4.18 J/(g·°C) (specific heat of water) ΔT = 100°C - 20°C = 80°C Solution: Q = mcΔT = 200 g × 4.18 J/(g·°C) × 80°C = 200 × 4.18 × 80 = 67,072 J Answer: Approximately 67,072 joules of heat are needed. Samp M Mazie Shields DVM Mar 28, 2026
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