Chapter 2 Energy and Matter

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