A saucepan contains 0.80 kg of water. It is heated from 25°C to 75°C. With c = 4184 J/kg°C, how much energy is added?

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Multiple Choice

A saucepan contains 0.80 kg of water. It is heated from 25°C to 75°C. With c = 4184 J/kg°C, how much energy is added?

When you heat a substance, the energy added (heat) is proportional to its mass, its specific heat capacity, and the temperature change: Q = m c ΔT. Here, mass is 0.80 kg, c is 4184 J/(kg·°C), and the temperature rises from 25°C to 75°C, so ΔT = 50°C. Multiply: Q = 0.80 × 4184 × 50 = 167,360 J. So about 1.67 × 10^5 joules of energy are added. This aligns with the idea that energy scales with mass, how much the temperature changes, and the material’s ability to store heat. The other numbers correspond to different temperature changes (for example, a 25°C rise would give 83,680 J, a 100°C rise would give 334,720 J), not the 50°C rise in this problem.

When you heat a substance, the energy added (heat) is proportional to its mass, its specific heat capacity, and the temperature change: Q = m c ΔT. Here, mass is 0.80 kg, c is 4184 J/(kg·°C), and the temperature rises from 25°C to 75°C, so ΔT = 50°C. Multiply: Q = 0.80 × 4184 × 50 = 167,360 J. So about 1.67 × 10^5 joules of energy are added. This aligns with the idea that energy scales with mass, how much the temperature changes, and the material’s ability to store heat. The other numbers correspond to different temperature changes (for example, a 25°C rise would give 83,680 J, a 100°C rise would give 334,720 J), not the 50°C rise in this problem.

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