Specific Heat Set

Introduction

The purpose of this activity is to demonstrate that five equal-massed metal specimens, when heated to the exact same temperature and added to a precise amount of water, will alter the temperature of the water to a significantly different extent. What is the reason for this? Each of the metals has a different specific heat!

Concepts

• Specific heat
• Heat capacity
• Calorimetry

Background

Transfer of heat or heat flow always occurs in one direction—from a region of higher temperature to a region of lower temperature—until some final equilibrium temperature is reached. The transfer of heat energy can be detected by measuring the resulting temperature change, ΔT, calculated by subtracting the initial temperature from the final temperature.

In this experiment, heat is transferred from a hot metal sample to a colder water sample. Each metal causes the temperature of water to increase to a different extent. This means that each metal must have a differing ability to absorb energy and release the energy to the water causing the temperature to rise. The ability of any material to contain heat energy is called that material’s heat capacity. The measure of heat capacity, or the quantity of heat needed to raise the temperature of one gram of a substance by one degree Celsius at constant pressure is termed specific heat, and is represented by the symbol C_p. The SI units for specific heat are given in J/g•°C and the non-SI units are cal/g•°C. (Note: 1 calorie = 4.184 Joules). The specific heats for the five metals included in this set are provided in Table 1.

Let’s compare the specific heats of aluminum and copper. Part of the reason for the variation of specific heats is that each substance is made up of atoms that have different masses. The mass of each copper atom is larger than the mass of each aluminum atom; therefore, a given mass (let’s say 58 g) of copper has fewer atoms than the same mass of aluminum. When heat is added to 58 g of copper, fewer atoms need to be put in motion. Thus, less heat is needed to increase the kinetic energy of the 58 g copper sample, and raise the temperature by 1 °C. Therefore, the specific heat of copper is lower than the specific heat of aluminum. Copper will also heat up more rapidly than aluminum, as it has less of an ability to “hold” heat before its temperature rises. Notice that copper and zinc have identical specific heat values. This is due to the similar mass of the atoms. In general, the larger the metal atom, the lower its specific heat.

Additional specific heat values are provided in Table 2. Compare the heat capacities of wood and the metals in general. Wood—with its relatively high specific heat value—is able to absorb more heat than metal before its temperature rises, and therefore does not feel hot to the touch. Metals, on the other hand, will heat up more quickly and feel hot to the touch due to their relatively low heat capacities. In general, the lower the specific heat value, the quicker the temperature will rise. Notice the high heat capacity of water. Water is able to absorb and store large amounts of heat. This has a moderating effect on air temperature near bodies of water. The amount of heat delivered by a material (q) is equal to the mass of the material delivering the heat (m) multiplied by the specific heat of the material (C_p) multiplied by the temperature change associated with delivering the heat (ΔT). The equation can be written as follows: To make accurate measurements of heat transfer and to prevent heat loss to the surroundings, an insulating device called a calorimeter is used. A calorimeter is a device used to measure heat flow, where the heat given off by a material is absorbed by the calorimeter and its contents (often water or other materials whose heat capacities are known).

The heat gained by the water in the calorimeter (or gained by the calorimeter itself if a dry calorimeter is used) must be equal in magnitude (and opposite in sign) to the heat lost by the sample, or Since then Equation 5 may be used to calculate the specific heat of an unknown metal sample.

Materials

Safety Precautions

Take precautions to avoid burns when heating the metal in the boiling water. Use tongs and allow the boiling water to cool before pouring it down the drain to prevent steam burns. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. Metal specimens should be dried and stored with other metals (Storage code—Inorganic #1).

Prelab Preparation

Procedure

1. Weigh a specific heat metal sample on a balance to the nearest tenth of a gram. Record this mass in a data table.
2. Place the metal sample in a boiling water bath for approximately 5–10 minutes to be sure the temperature of the sample is 100 °C.
3. Fill a calorimeter with a measured quantity of room temperature or slightly chilled water. Record the mass of the water used. Measure and record the temperature of the water in the calorimeter in °C.
4. Using tongs, lift up the heated metal sample from the boiling water bath and carefully place it into the water in the calorimeter.
5. Stir the water in the calorimeter with a stirring rod, slowly and constantly. Use a thermometer to measure and record the highest temperature that the water reaches. Note: Remember that the heat gained by the water equals the heat lost by the sample.

Calculations (if using a known metal of known specific heat) 

6. Calculate the heat gained by the water using the mass of the water, the temperature change of the water in °C and the specific heat of the water. Hint: Use Equation 2.
7. Using Equation 3, calculate the heat lost by the metal based on the mass of the sample, the temperature change of the metal (recall its initial temperature was 100 °C), and the specific heat of the metal. Compare the values of the heat gained by the water and the heat lost by the metal. How do they compare? If the values are not equal, where might the missing heat have gone?

Calculations (if using a known metal with unknown specific heat)

8. Calculate the heat gained by the water using the mass of the water, the temperature change of the water in °C and the specific heat of the water. Hint: Use Equation 2.
9. Calculate the specific heat of the metal using the mass of the sample, the temperature change of the metal (Note: Initial temperature was 100 °C.) and the heat gained by the water (from Question 8). Hint: Use Equation 5.

Calculations (if using an unknown metal)

10. Calculate the heat gained by the water using the mass of the water, the temperature change of the water in °C and the specific heat of the water. Hint: Use Equation 2.
11. Determine the identity of the unknown metal by calculating its specific heat according to step 9 and comparing the value to a list of published values.

Teacher Tips

• This handout is written as a teacher reference and is not specifically intended for student copy purposes. Purchase of this specific heat set, however, provides you with permission to use any part of this handout in the manner desired.
• This lab activity works well using a coffee cup calorimeter, which is made by simply placing two insulated foam coffee cups together (for extra insulation) and filling the cup with the measured amount of water. Caution students not to leave thermometers standing up in any cup as the cup may tip over and the thermometer may break.
• Several of the metals are similar in color and size. The following descriptions may help distinguish and identify tin, zinc and steel.

  Zinc: Dull, silver gray, surface is slightly cratered or pockmarked.
  Tin: Light silver, shiny and reflective, pockmarked.
  Steel: Silver or gray, polished or smooth metal surface.

Answers to Questions

Solution to Sample Problem

1. Using Equation 2,

   q (gained by water) = m (water) x C_p (water) x ΔT (water)
   q (gained by water) = (60 g) x (1.00 cal/g °C) x (4 °C) = 240 calories

2. Using Equation 5,

   m (water) x C_p (water) x ΔT (water) = –[m (metal) x C_p (metal) x ΔT (metal)],
   (60 g) x (1.00 cal/g °C) x (4 °C) = –[(58 g) x °C_p) x (–78 °C)]
   C_p = 0.053 cal/g•°C, so using the table of specific heats, the metal is tin.

Additional Problems

1. How much energy (in Joules) is needed to heat an iron nail with a mass of 7.0 grams from 25 °C until it becomes red hot at 750 °C?

   Solution: q = (7.0 g) x (0.448 J/g °C) x (725 °C) = 2274 joules of energy

2. How much energy (in calories) is in a large peanut that heats up 100 g of water from 20 °C to 65 °C? (Note: Food calories are given in kilocalories where 1 kcal = 1000 cal)

   Solution: q = (100 g) x (1.00 cal/g °C) × (45 °C) x (1 kcal/1000 cal) = 4.5 kcal

Discussion

Sample Problem
If a 58-g sample of metal at 100 °C is placed into a calorimeter containing 60 g of water at 18 °C, the temperature of the water increases to 22 °C.

1. Calculate the amount of heat transferred to the water.
2. Determine the identity of the metal by calculating its specific heat. Note: Assume no heat loss to the surroundings.

Additional Problems

1. How much energy (in Joules) is needed to heat an iron nail with a mass of 7.0 grams from 25 °C until it becomes red hot at 750 °C?
2. How much energy (in calories) is in a large peanut that heats up 100 g of water from 20 °C to 65 °C? (Note: Food calories are given in kilocalories where 1 kcal = 1000 cal)