A 100 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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Prepare for the Dual Enrollment Physical Science Midterm. Study with flashcards and multiple choice questions, each with hints and explanations. Ace your exam!

Multiple Choice

A 100 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:
Heating a substance uses the relation q = m c ΔT, which gives the energy required to raise its temperature by ΔT for a given mass and specific heat. Here, ΔT = 80°C − 20°C = 60°C. The mass is 100 g and the specific heat of water is 4.18 J/g°C. So q = 100 g × 4.18 J/g°C × 60°C = (4.18 × 60) × 100 = 250.8 × 100 = 25,080 J. Therefore, the energy required is 25,080 joules (about 25 kJ).

Heating a substance uses the relation q = m c ΔT, which gives the energy required to raise its temperature by ΔT for a given mass and specific heat. Here, ΔT = 80°C − 20°C = 60°C. The mass is 100 g and the specific heat of water is 4.18 J/g°C. So q = 100 g × 4.18 J/g°C × 60°C = (4.18 × 60) × 100 = 250.8 × 100 = 25,080 J. Therefore, the energy required is 25,080 joules (about 25 kJ).

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