Specific Heat and Calorimetry - Chemistry

1. Fundamental Concepts

Energy: The capacity to do work or produce heat. In this context, we focus on Heat (), the energy transferred due to a temperature difference.
Temperature: A measure of the average kinetic energy of the particles in a substance.
Specific Heat Capacity (): The amount of heat required to raise the temperature of 1 gram (1g) of a substance by 1°C (or 1K).
- Units: or .
- Water's specific heat: (Memorize this value).
Calorimetry: The experimental technique used to measure the heat absorbed or released during a physical or chemical process.
Calorimeter: The insulated device used to measure these heat changes (e.g., Coffee Cup Calorimeter).

2. Key Concepts

The Specific Heat Formula:

$$q = mc\Delta T$$

$q$: Heat energy (Joules)

$m$: Mass (grams)

$c$: Specific heat capacity

$\Delta T$: Change in temperature ($T_{final} - T_{initial}$)

Sign Convention:

$q > 0$ (Positive): Heat is absorbed (Endothermic). Temperature increases.

$q < 0$ (Negative): Heat is released (Exothermic). Temperature decreases.

Law of Conservation of Energy:

In an insulated system, the heat lost by one substance equals the heat gained by another.

$$q_{system} + q_{surroundings} = 0$$

Commonly expressed as: $q_{metal} + q_{water} = 0$ or $-q_{lost} = q_{gained}$

3. Examples

Easy - Basic Formula Application

Question: Calculate the heat required to warm 50.0g of water from 20.0°C to 100.0°C? (Water's $c = 4.18 J/g°C$)

Solution:

1. $\Delta T = 100.0 - 20.0 = 80.0°C$

2. $q = (50.0g)(4.18 J/g°C)(80.0°C)$

3. $q = 16,720 J$ or $16.7 kJ$

Medium - Mixing Substances (Finding Final Temp)

Question: 100.0g of water at 25.0°C is mixed with 50.0g of water at 80.0°C. What is the final temperature ($T_f$)? (Assume no heat loss to the surroundings and round to one decimal places).

Solution:

1. Heat gained by cold water = - Heat lost by hot water

2. $m_1c(T_f - T_1) = -m_2c(T_f - T_2)$ (The $c$ cancels out).

3. $100(T_f - 25) = -50(T_f - 80)$

4. $100T_f - 2500 = -50T_f + 4000$

5. $150T_f = 6500 \Rightarrow T_f \approx 43.3°C$

Hard - Calorimetry (Finding Specific Heat)

Question: A 55.0g block of an unknown metal at 99.8°C is placed into 100.0g of water at 22.0°C. The final temperature of the mixture is 28.5°C. Calculate the specific heat ($c$) of the metal. (Round to three decimal places; Water's $c = 4.18 J/g°C$)

Solution:

1. $q_{metal} + q_{water} = 0$

2. $(55.0)(c_{metal})(28.5 - 99.8) + (100.0)(4.18)(28.5 - 22.0) = 0$

3. $(55.0)(c_{metal})(-71.3) + (100)(4.18)(6.5) = 0$

4. $-3921.5 c_{metal} + 2717 = 0$

5. $3921.5 c_{metal} = 2717$

6. $c_{metal} \approx 0.693 J/g°C$

4. Problem-Solving Techniques

Organize Data: Create a list of knowns and unknowns ($m, c, T_i, T_f$) for both the system (e.g., metal) and the surroundings (e.g., water).

Check Units: Ensure mass is in grams and energy units are consistent (Joules or calories).

Calculate $\Delta T$ Correctly: Always use $\Delta T = T_{final} - T_{initial}$. If the object is cooling down, $\Delta T$ will be negative, making $q$ negative.

The "One Equation" Strategy:

Instead of calculating $q$ for each side separately and guessing signs, use the conservation of energy equation directly:

$$m_1c_1(T_f - T_1) + m_2c_2(T_f - T_2) = 0$$

Plug in the numbers and solve for the unknown.

Estimate $T_f$: The final temperature must always be between the two initial temperatures. If your calculation gives a $T_f$ higher than the hottest substance or lower than the coldest, you have an error.