Heat Capacity and Specific Heat in GOB Chemistry

Heat–Temperature Relationship

Introduction to Heat and Temperature

When an object is heated, its temperature increases because heat is directly proportional to its temperature change. The relationship is expressed as:

• Heat (q) is the energy transferred due to temperature difference.

• Temperature change (ΔT) is the difference between final and initial temperature.

Equation:

$q \propto \Delta T$

Key Point

• As you heat an object, the amount of heat required increases with the temperature change.

Example

• If the temperature of a water bath goes from 25 K to 50 K, the amount of heat will double.

Molar & Specific Heat Capacity

Definitions and Concepts

Heat Capacity (C): The amount of heat required to change the temperature of a weighted substance.

• Specific Heat Capacity (c): Amount of heat required to change the temperature of 1 g of substance by 1 K.

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

Formulas

Molar Heat Capacity	Specific Heat Capacity
$C_m = \frac{q}{n \Delta T}$ q = heat (J) n = moles ΔT = temperature change (°C)	$c = \frac{q}{m \Delta T}$ q = heat (J) m = grams ΔT = temperature change (°C)

Example

• If 15.7 g of silver raises its temperature by 17.2 °C when it absorbs 6845.5 J, its molar heat capacity is:

$\frac{6845.5}{15.7 \times 17.2} = 25.4 \text{ J/g·°C}$

$273.1 \times 10^{-3} \text{ J/mol·°C}$

Specific Heat Capacity Formula

Calculating Heat Transfer

The specific heat capacity formula allows calculation of heat released or absorbed:

$q = mc\Delta T$

• q = heat (J)

• m = mass (g)

• c = specific heat capacity (J/g·°C)

• ΔT = temperature change (°C)

Example

• How much heat (in kJ) is released when 120.0 g H2O goes from 90 °C to 45 °C? c = 4.184 J/g·°C

$q = 120.0 \times 4.184 \times (45 - 90)$ $= 120.0 \times 4.184 \times (-45)$ $= -22,593.6 \text{ J} = -22.6 \text{ kJ}$

Practice Problems and Applications

Comparing Temperature Changes

• Given specific heat capacities, the substance with the lowest specific heat will show the greatest temperature change for a given amount of absorbed heat.

Practice Example

• Which compound will show the greatest temperature change upon absorbing 250 J of heat?

• Options: 250.0 g Al, 250.0 g Cu, 250.0 g ethanol, 250.0 g wood

• Answer: 250.0 g Cu (lowest specific heat capacity)

Specific Heat Capacities Table

Substance	Specific Heat Capacity (J/g·°C)
Aluminum, Al	0.900
Copper, Cu	0.385
Ethanol, C2H5OH	2.450
Water, H2O	4.184
Wood	1.760

Additional Practice

• A sample of copper absorbs 3.53 kJ of heat, increasing temperature by 25 °C. Find the mass (kg) if c = 0.385 J/g·°C.

$q = mc\Delta T$ $3530 = m \times 0.385 \times 25$ $m = \frac{3530}{0.385 \times 25} = 368.8 \text{ g} = 0.369 \text{ kg}$

• 50.00 g of heated metal is placed into 822.5 g water at 32.08 °C. If the metal gains 14.55 kJ, what is the final temperature of the water? c (water) = 4.184 J/g·°C.

$q = mc\Delta T$ $14,550 = 822.5 \times 4.184 \times \Delta T$ $\Delta T = \frac{14,550}{822.5 \times 4.184} = 4.29 \text{°C}$ Final temperature = 32.08 + 4.29 = 36.37 °C

Summary Table: Key Equations

Equation	Variables	Application
$q = mc\Delta T$	q = heat, m = mass, c = specific heat, ΔT = temp change	Calculating heat for temperature change
$C_m = \frac{q}{n\Delta T}$	q = heat, n = moles, ΔT = temp change	Molar heat capacity
$c = \frac{q}{m\Delta T}$	q = heat, m = mass, ΔT = temp change	Specific heat capacity

Additional info: These notes cover the essential concepts of heat capacity, specific heat, and their calculations, which are fundamental topics in GOB Chemistry. Practice problems and tables are included to reinforce understanding and application.