Heat Capacity and Specific Heat: Study Notes for Introduction to Chemistry

Heat Capacity and Temperature Change

Concept of Heat Capacity

Heat capacity is a fundamental property in chemistry that describes how much heat energy is required to change the temperature of a substance. The temperature increase of an object is directly proportional to the amount of heat added.

• Heat-Temperature Relationship: The relationship between heat (q), mass (m), specific heat capacity (c), and temperature change (ΔT) is given by:

$q = mc\Delta T$

• Example: If the temperature of a water bath goes from 25°C to 50°C, the amount of heat required will double if the mass is doubled, and will triple if the temperature change is tripled.

Molar and Specific Heat Capacity

Heat capacity can be expressed in two main ways: molar heat capacity and specific heat capacity.

• Molar Heat Capacity (Cm): The amount of heat required to change the temperature of 1 mole of a substance by 1 K.

• Specific Heat Capacity (c): The amount of heat required to change the temperature of 1 gram of a substance by 1 K.

Molar Heat Capacity	Specific Heat Capacity
$C_m = \frac{q}{n\Delta T}$ where $q$ = heat (J), $n$ = moles, $\Delta T$ = temperature change (K or °C)	$c = \frac{q}{m\Delta T}$ where $q$ = heat (J), $m$ = mass (g), $\Delta T$ = temperature change (K or °C)

• Example: If 15.0 g of silver raises its temperature by 17.2°C when it absorbs 648.5 J, its molar heat capacity can be calculated using the above formula.

Specific Heat Capacity Formula

Measurement and Calculation

The specific heat capacity formula allows us to calculate the amount of heat released or absorbed by a substance during a temperature change.

$q = mc\Delta T$

• Example: How much heat (q) is released when 120 g H2O cools from 95°C to 45°C? The specific heat capacity of H2O is 4.184 J/g·°C.

• Calculation: $q = 120 \times 4.184 \times (45 - 95) = -25,104$ J (heat released)

Practice Problems

• Example: A sample of copper absorbs 5.13 kJ of heat, increasing its temperature by 25°C. Find the mass of copper if its specific heat capacity is 0.385 J/g·°C.

• Calculation: $m = \frac{q}{c\Delta T} = \frac{5130}{0.385 \times 25} = 533$ g = 0.73 kg

Comparing Specific Heat Capacities

Temperature Change and Substance Type

Different substances have different specific heat capacities, which affect how much their temperature changes when they absorb the same amount of heat.

Substance	Specific Heat Capacity (J/g·°C)
Silver (Ag)	0.235
Water (H2O)	4.184
Ethanol	2.44
Wood	1.76

• Practice: Based on their specific heat capacities, silver will show the greatest temperature change upon absorbing 25.0 J of heat, because it has the lowest specific heat capacity.

Application Example

• Example: 50.0 g of heated metal is placed in 62.5 g of water. If the metal gives 14.5 J of energy to the water, and the final temperature of the water is 20.0°C, the initial temperature of the water can be calculated using the specific heat capacity formula.

• Calculation: $\Delta T = \frac{q}{mc} = \frac{14.5}{62.5 \times 4.184} = 0.056$°C; Initial temperature = 20.0°C - 0.056°C = 19.94°C

Additional info: These notes cover the topic of heat capacity, specific heat, and their application in temperature change calculations, which are part of Chapter 3 (Matter and Energy) and Chapter 6 (Chemical Composition) in an Introduction to Chemistry course.