Specific Heat and Calorimetry

Specific Heat and Calorimetry Problems

Introduction to Specific Heat Equation

• 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.

Coffee Cup Calorimetry

• 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.

Energy Transfer

• 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).

Calorimeters

• 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.

Application of Calorimetry

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

Labs

• 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.

Mixing Hot and Cold Water

• 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 $\Delta T = T<em>{final} - T</em>{initial}$ and $c$ is specific heat.

• $m<em>1 \cdot (T</em>{final} - T<em>{initial(1)}) \cdot c = -[m</em>2 \cdot (T<em>{final} - T</em>{initial(2)}) \cdot c]$"

  • 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.

Neutralization Reaction and Calorimetry

• Problem 11 involves the neutralization of $KOH$ with sulfuric acid.

• Given: 1 mole of $KOH$ neutralized releases -56 kJ.

• Balanced equation:

  $2KOH + H<em>2SO</em>4 \rightarrow K<em>2SO</em>4 + 2H_2O$ with $\Delta H = -112 kJ$

• Initial conditions: 25.0 mL of 0.475 M $H<em>2SO</em>4$ at $23.7 °C$ neutralized by 0.613 M $KOH$.

• Goal: Find the final temperature of the solution after neutralization.

• First, use $M<em>1V</em>1 = M<em>2V</em>2$ (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.