## Lab 9 Introduction

### Learning Objectives

By the end of this section, you will be able to:

• Determine whether a reaction is endothermic or exothermic.
• Determine the best ionic compound of to use in a heat pack for treating frostbite based on your experimental results.
• Calculate the average heat capacity of your calorimeter.
• Calculate the specific heat of a metal.

## Introduction

The human body works best within a very narrow temperature range. A temperature drop of as little as 2$^{\circ}$C in the body’s core causes classic hypothermia symptoms such as mental difficulties and loss of physical coordination. Much more extreme temperature drops in the extremities may be survived, but can lead to frostbite if the flesh freezes. Victims of hypothermia require immediate treatment, and in outdoor situations the treatment is often warmth provided by portable heat sources, such as heat packs.

Heat packs are available that produce warmth through various chemical reactions. Such heat packs are convenient because they only release heat when triggered. One common heat pack contains an internal pouch of water and a solid powder. Once the pouch of water is broken open, there is an exothermic reaction between the water and the powder. These heat packs have limitations. For example, they do not work well in extreme cold because the water within the pack will freeze.

In the coldest environments, heat packs are available that contain only the powder in a resealable waterproof sack. When heat is required, the sack can be opened and any aqueous solution poured inside. The sack is resealed and the reaction produces heat.

### Reaction Enthalpy

An important part of chemistry involves studying energy changes that occur during chemical reactions. These energy changes are of fundamental importance in understanding the “driving force” of a chemical reaction. The most common way energy is exchanged between a chemical system and the environment is by evolution or absorption of heat (q). The change in heat energy accompanying a chemical reaction is known as enthalpy change, ΔH. By convention, reactions in which heat is absorbed are labeled endothermic and have positive values of ΔH; reactions in which heat is released are labeled exothermic and have negative values of ΔH.

Cold packs used in athletics are familiar to many sports enthusiasts. In order to derive coldness from the pack, a plastic packet of water is broken inside another packet containing a solid salt such as NH4NO3. In this case, the enthalpy of solution, i.e., heat absorbed when a substance dissolves, is endothermic indicating that heat is absorbed as the salt dissolves. Thus, the enthalpy of solution is designated with a positive sign since energy is absorbed or added.

$\text{NH}_{4}\text{NO}_{3}\text{(s)}\rightarrow\text{NH}_{4}\text{NO}_{3}\,\,\,\,\,\,\,\Delta\text{H}=+25.7\text{ kJ}$

On the other hand, gas stoves produce heat through the combustion (burning in oxygen) of methane, equation 1. Since heat is produced by the combustion, the reaction is exothermic and enthalpy of combustion (ΔH), i.e., heat released during combustion, must be negative. In fact, ΔH for the methane combustion in is –890.4 kJ (see below). It is important to realize that ΔH is related to the coefficients in the balanced equation. Thus, –890.4 kJ of heat is released per every 1 mole of CH4 that reacts but per every two moles of H2O that is formed.

Equation 1

$\text{CH}_{4(g)}+\text{O}_{2(g)} →\text{CO}_{2(g)}+\text{2H}_{3(g)}\text{O}\,\,\,\,\,\,\,\Delta\text{H}=-890.4\text{ kJ}$

The negative sign serves to reinforce the fact that heat is produced rather than absorbed or that this is an exothermic reaction. We can use stoichiometry to calculate the enthalpy of any amount of a reactant.

### Heat and Temperature

If an object (such as a pot of water) is positioned to absorb the heat given off during a combustion reaction, then the temperature of the object will change as follows

Equation 2

$\text{q}=\text{mc}\Delta\text{T}$

where

 q = the amount of heat absorbed by the object in Joules m = the mass of the object being heated in grams c = the specific heat of the object being heated ΔT = the change in temperature of the object = final   temperature minus the initial temperature = TF – TI

The specific heat is different for different substances. The specific heat or heat capacity is low for objects which are easily heated or cooled with minimal energy input. Metals usually have low specific heats because their temperature changes very quickly. Water has a high specific heat because it takes a lot of energy to change the temperature of a sample of water. For example:

 Substance Specific Heat (J g-1 $^{\circ}$C-1) water 4.18 air 1.01 aluminum 0.897 granite 0.790

## Calorimetry

Many experiments in thermochemistry involve a calorimeter. A calorimeter, like the one to the right, is simply a container that insulates a reaction from the surrounding environment so change in temperature as the reaction proceeds place can be measured accurately measured as independent of the environmental temperature. In lab, you will use two styrofoam cups as your calorimeter.

Ideally any calorimeter would be able to completely maintain the system without losing heat to the surroundings. Unfortunately, no calorimeter is perfect and heat is always lost to the surroundings. It is possible to determine how much heat is lost by the calorimeter by mixing hot water and room temperature water. The heat lost by the hot water is equal to the heat gained by the room temperature water.

Equation 3

$|q_{\text{lost by hot water}}|=|q_{\text{gained by room temperature water}}|+|q_{\text{gained by calorimeter}}|$

In order to calculate the heat lost to the calorimeter you will calculate q for both the hot water and the room temperature water (using Equation 2) and then subtract to find the difference.

It is also possible to view how heat can be transferred from one object to another. In a perfect system (where no heat is lost to the calorimeter) the heat lost by the hot object would be gained by the cold object such as

Equation 4

which could be rearranged to give

$q_{\text{lost}} = -q_{\text{gained}}$

Equation 5

as well as replacing q with Equation 3 to give

$q_{\text{hot water}}=-q_{\text{cold water}}$

Equation 6

$\text{mc}{\text({T}_{f} - {T}_{i}})_\text{hot object}=-\text{mc}{\text({T}_{f} - {T}_{i}})_\text{cold object}$

Although usually heat is lost to the calorimeter, as we showed in Equation 3, we can usually assume this value is exceptionally small and omit it from the equation as we did in Equations 4-6.

In this lab, we will be investigating the endothermic and exothermic qualities of salt solutions by dissolving various salts into water and monitoring the temperature. Then you will calibrate a coffee cup calorimeter by determining its heat capacity. Finally, you will calculate the specific heat of copper by dropping hot pennies into water and monitoring the heat exchange.