Specific Heat and Calorimetry

Specific Heat and Calorimetry Problems

The lecture introduces the topic of specific heat using problems.
The lab involves finding the specific heat of an unknown metal object using a coffee cup calorimeter (a styrofoam cup).
Picking up metal from boiling water with bare hands is not recommended.

A coffee cup calorimeter is used to measure the energy given off or absorbed in a reaction happening in water.
By knowing the mass of water in the cup, the initial temperature, adding a hot piece of metal, and measuring the final temperature, one can calculate the energy exchange.
The specific heat of water is always known: 4.18 \frac{J}{g \cdot °C}.
Everything is measured relative to water.

When a hot piece of metal is placed in water, the energy the metal loses as it cools is gained by the water.
Chemical reactions in water can also be measured this way.
Example: Burning a peanut under a container of water to measure the energy released (though this is not done anymore due to allergies).

Real calorimeters are well-insulated containers filled with a known quantity of water at a known temperature.
A reaction happens inside a chamber within the calorimeter.
The mass of all materials (metal, water, etc.) and their specific heats are known.
Water is used as a heat sink because it absorbs a lot of heat without drastically changing temperature.

Calorimeters are used to measure the energy released by burning foods to determine calorie content for nutrition labels.

Labs will involve determining the specific heat of different metals and the heat of fusion of ice.
Experimental results usually come within a few percent of the actual values.
Metals are heated in boiling water because the temperature is known (100 °C). It's hard to measure temperature in an open flame.

Problem: Mixing 375 mL of water at 32 °C with 25 mL of water at 92 °C. Find the final temperature.
Energy lost by hot water equals the energy gained by cold water.
Equation: q = m \cdot c \cdot \Delta T where \Delta T = T{final} - T{initial} and c is specific heat.
m1 \cdot (T{final} - T{initial(1)}) \cdot c = -[m2 \cdot (T{final} - T{initial(2)}) \cdot c]"
- Since both are water, specific heat (c) cancels out.
- Density of water is approximately 1 \frac{g}{mL}, so volumes can be converted to masses.
- Solve for T_{final}, which is the same for both cups after mixing.
- The negative sign accounts for the hot water losing energy (cooling off), making its \Delta T negative.

Problem 11 involves the neutralization of KOH with sulfuric acid.
Given: 1 mole of KOH neutralized releases -56 kJ.
Balanced equation:
2KOH + H2SO4 \rightarrow K2SO4 + 2H_2O with \Delta H = -112 kJ
Initial conditions: 25.0 mL of 0.475 M H2SO4 at 23.7 °C neutralized by 0.613 M KOH.
Goal: Find the final temperature of the solution after neutralization.
First, use M1V1 = M2V2 (accounting for stoichiometry) to find the volume of KOH needed.
- V_{KOH} = 38.7 mL
Total volume = 25.0 mL + 38.7 mL = 63.7 mL, which equals 63.7 grams (since density is 1 g/mL).
Calculate moles of KOH using molarity: Molarity = \frac{moles}{Liters}, so moles = 0.613 M \cdot 0.0387 L = 0.024 moles.
Calculate energy released: 0.024 moles KOH \cdot \frac{-56 kJ}{1 mole KOH} = -1.344 kJ = -1344 J.
Use q = m \cdot c \cdot \Delta T to find the final temperature.
-1344 J = 63.7 g \cdot 4.18 \frac{J}{g \cdot °C} \cdot (T_{final} - 23.7 °C)
The energy released by the reaction is absorbed by the water, causing the temperature to rise.