- Problem 1: The specific heat of iron is calculated using C = q / (m × ΔT). With q = 1086.75 J, m = 16.03 g, and ΔT = 175°C - 25°C = 150°C, the calculation is C = 1086.75 J / (16.03 g × 150°C) = 0.45 J/g·°C.
- Problem 2: The heat required is calculated using q = m × C × ΔT. With m = 47.5 g, C = 0.90 J/g·°C, and ΔT = 94°C - 21°C = 73°C, the calculation is q = 47.5 g × 0.90 J/g·°C × 73°C = 3100 J.
- Problem 3: The specific heat capacity of wood is calculated using C = q / (m × ΔT). With q = 67,500 J, m = 1500.0 g, and ΔT = 57°C - 32°C = 25°C, the calculation is C = 67,500 J / (1500.0 g × 25°C) = 1.8 J/g·°C.
- Problem 4: The heat energy needed for water is calculated using q = m × C × ΔT. With m = 348 g, C = 4.18 J/g·°C (standard value for water), and ΔT = 37°C - 4.0°C = 33°C, the calculation is q = 348 g × 4.18 J/g·°C × 33°C = 48,000 J.
Parent Tip: Review the logic above to help your child master the concept of heat of fusion worksheet.
