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Understanding Specific Heat: Key Concepts and Calculations in Energy Transfer

This chapter explores the concept of specific heat, which quantifies the heat required to raise the temperature of 1 gram of a substance by 1°C. It highlights different materials' specific heat values, the difference between SI and metric units, and the importance of specific heat in everyday situations such as ocean water cooling and sandy desert conditions. Additionally, practical learning checks are provided to reinforce the application of the heat equation in real-world scenarios, showcasing how to calculate the energy required to change temperatures.

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Understanding Specific Heat: Key Concepts and Calculations in Energy Transfer

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  1. Chapter 2 Energy and Matter 2.4 Specific Heat

  2. Specific Heat Specific heat • is different for different substances. • is the amount of heat that raises the temperature of 1 g of a substance by 1°C. • in the SI system has units of J/gC. • in the metric system has units of cal/gC.

  3. Examples of Specific Heats TABLE 2.6 cal/g°C 0.214 0.0920 0.0308 0.108 0.0562 0.125 0.488 0.588 0.207 0.100

  4. Learning Check A. When ocean water cools, the surrounding air 1) cools. 2) warms. 3) stays the same. B. Sand in the desert is hot in the day, and cool at night. Sand must have a 1) high specific heat. 2) low specific heat.

  5. Heat Equation Rearranging the specific heat expression gives the heat equation. Heat = g x °C x J = J g°C The amount of heat lost or gained by a substance is calculated from the • mass of substance (g). • temperature change (T). • specific heat of the substance (J/g°C).

  6. Solution A. When ocean water cools, the surrounding air 2) warms. B. Sand in the desert is hot in the day, and cool at night. Sand must have a 2) low specific heat.

  7. Learning Check What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2C to 24.5C?

  8. Solution What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2C to 24.5C? Given: 24.8 g metal, 275 J of energy, 20.2C to 24.5C Need: J/gC Plan: SH = Heat/gC ΔT = 24.5C – 20.2C = 4.3 C Set Up: 275 J = 2.6 J/gC (24.8 g)(4.3C)

  9. Learning Check How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C? 1) 20.4 kJ 2) 77.7 kJ 3) 84.3 kJ

  10. Solution How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 3) 84.3 kJ 77.0°C – 15.0°C = 62.0°C 325 g x 62.0°C x 4.184 J x 1 kJ g °C 1000 J = 84.3 kJ

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