Calculate a heat-capacity model

Supplied materials

• All measurements and observations below are supplied classroom examples. They are not data from actual learners or a claim about a particular real site.
• Paper or browser document; an ordinary calculator is optional.
• Rule: Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. Formula: a*b*c. Variable key: a = mass (kg), b = specific heat (J/(kg·K)), c = temperature increase (K)
• Distinct worked inputs and result: A distinct model has 2 kg, specific heat 150 J/(kg·K) and change 4 K: 2×150×4=1,200 J.
• Assessment case 1: Supplied case 1: mass (kg)=2; specific heat (J/(kg·K))=100; temperature increase (K)=3, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?
• Assessment case 2: Supplied case 2: mass (kg)=3; specific heat (J/(kg·K))=200; temperature increase (K)=2, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?
• Assessment case 3: Supplied case 3: mass (kg)=1.5; specific heat (J/(kg·K))=400; temperature increase (K)=4, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?
• Assessment case 4: Supplied case 4: mass (kg)=4; specific heat (J/(kg·K))=250; temperature increase (K)=2, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?
• Assessment case 5: Supplied case 5: mass (kg)=2.5; specific heat (J/(kg·K))=120; temperature increase (K)=5, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?
• Assessment case 6: Supplied case 6: mass (kg)=5; specific heat (J/(kg·K))=80; temperature increase (K)=3, using the explicitly named quantities. Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. What result follows?

Learn the idea

For a sample with constant specific heat and no phase change, energy transfer can be modeled by mass times specific heat times temperature change. Real materials and phase transitions can require more detail.

Energy transferred equals kilograms times supplied joules per kilogram-kelvin times temperature change in kelvin. The formula in the supplied model is a*b*c. Variable key: a = mass (kg), b = specific heat (J/(kg·K)), c = temperature increase (K). Use mass in kilograms for this supplied heat-capacity unit. A Celsius interval equals the same kelvin interval.

A distinct model has 2 kg, specific heat 150 J/(kg·K) and change 4 K: 2×150×4=1,200 J. The model excludes a phase change. Track the quantity and unit before calculating. Explain the meaning of a positive, negative or fractional result when relevant. A numerical answer supports the stated model calculation; broader claims require the additional measurements and assumptions described here.

1. Name the measured quantities and their units before substituting values.
2. Write the stated rule and use the variable key to match each symbol to its explicitly named quantity.
3. Calculate each supplied case separately, keeping intermediate arithmetic and the result unit.
4. Explain the result in context and identify the assumption that would invalidate direct use of the rule.

Check your reasoning

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Take it further

Changed-input task: use the same stated model with mass (kg)=3; specific heat (J/(kg·K))=100; temperature increase (K)=3. Expected result: 900 J. Explain what changes and keep this assumption: The model excludes a phase change.

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