A 50 g sample of water is heated from 20°C to 80°C. Using c = 4.18 J/g°C, how much energy is required?

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Multiple Choice

A 50 g sample of water is heated from 20°C to 80°C. Using c = 4.18 J/g°C, how much energy is required?

Explanation:
Calculating energy to heat a substance uses q = m × c × ΔT. Here, the mass is 50 g, the specific heat of water is 4.18 J/g°C, and the temperature change is from 20°C to 80°C, so ΔT = 60°C. Compute: 50 × 4.18 × 60 = (4.18 × 60) × 50 = 250.8 × 50 = 12,540 J. The energy required is 12,540 joules (about 12.54 kJ). This reflects how heating water needs a lot of energy per degree because of its relatively high specific heat.

Calculating energy to heat a substance uses q = m × c × ΔT. Here, the mass is 50 g, the specific heat of water is 4.18 J/g°C, and the temperature change is from 20°C to 80°C, so ΔT = 60°C. Compute: 50 × 4.18 × 60 = (4.18 × 60) × 50 = 250.8 × 50 = 12,540 J. The energy required is 12,540 joules (about 12.54 kJ). This reflects how heating water needs a lot of energy per degree because of its relatively high specific heat.

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