Lab 10 Lab Report

.pdf

School

Barry University *

*We aren’t endorsed by this school

Course

151

Subject

Chemistry

Date

Dec 6, 2023

Type

pdf

Pages

4

Uploaded by DeanFly2674

Report
1 m Lab 10 “Heat Transfer.” Heat is the amount of energy transferred from one high temperature object to a low temperature object. Heat transfer leads to change in temperature of the objects, it drops the temperature of high temperature object and increases that of the low temperature object. This process leads to thermal equilibrium, when the objects have the same temperature. The temperature change of a material depends on their Specific Heat Capacity (c), the amount of energy required to change the temperature of 1? of material, 1 ° 𝐶 . For example, ? 𝑤???? = 4.186 𝐽 = 1 ?𝑎? . The Calorie content in foods are based on 1 𝐶𝑎? = 1000 ?𝑎?. A calorimeter is an apparatus used to study changes in temperature. The calorimeter design permits to isolate the system being studied, heat cannot enter or leave the system from the outside. A thermos, lunch box, or styrofoam cup are all examples of imperfect calorimeters (a sealed styrofoam cup is actually a very good calorimeter, styrofoam is a very good thermal insulator, it is composed of many trap air bubble). In a perfect calorimeter, no energy leaves or enters the system from the outside. In this lab, you will use a calorimeter and the specific heat of water to compare the heat loss of hot water to the heat gain of cold water. The specific heat of a non-perfect calorimeter will be accounted for in the calculations using the specific heat of the metal it is made of. Materials: Calorimeter Thermometers Ice water Beaker Hot water Electronic Balanced 100 mL Graduated Cylinder PROCEDURE 1. Specific Heat Capacity of water is ? 𝑤𝑎??? = 1 ?𝑎? 2. Specific Heat Capacity of Aluminum ? 𝐴? = 0.22 ?𝑎? 3. Determine the mass of the calorimeter cup: ? ??? = ? 58.15 Part A: Adding Hot water to Cold water. ? ???+???? 𝑤𝑎??? = ? 157.35 : ? ???? 𝑤𝑎??? = 99.22 : ? ???𝑎? = ? 278.27 : ? ℎ?? 𝑤𝑎??? = 120.92 ? TABLE I: 𝑀𝑎?? (?) Initial Temperature 𝑇 𝑖 (℃) Final Temperature 𝑇 ? (℃) Heat 𝑄(?𝑎?) Cold Water 129.07 g 8.7 °C 43.6 °C 4508.83 Q Calorimeter Cup 58.15 g 8.7 °C 43.6 °C 557.77 Q
2 Hot Water 95.41 g 82 °C 43.6 °C 5272.11 Q Analysis: The heat gained by the cold water and the calorimeter cup is equal to the heat lost by the hot water, if any other transfer of heat is small enough to be neglected (for example, to the air). The amount of heat gained or lost by a substance without changing states is given by: 𝐻?𝑎? = ?𝑎?? ∙ ?𝑝??𝑖?𝑖? ℎ?𝑎? ?𝑎𝑝𝑎?𝑖?𝑦 ∙ ?ℎ𝑎??? 𝑖? ???𝑝??𝑎???? 𝑄 = ? ∙ ? ∙ Δ𝑇 Did cold water gain or lose heat? How did you know? Calculate the amount, record it in table I. 𝑄 ???? 𝑤𝑎??? = ? ???? 𝑤𝑎??? ∙ ? 𝑤𝑎??? ∙ Δ𝑇 = 99.2g x 1 cal x 43.6 °C= 4325.12 Q Did calorimeter cup gain or lose heat? How did you know? Calculate the amount, record it in table I. 𝑄 ?𝑎???𝑖????? = ? ?𝑎???𝑖????? ∙ ? 𝐴? ∙ Δ𝑇 58.15 g x 0.22 cal x 43.6 °C =557.77 Q Did hot water gain or lose heat? How did you know? Calculate the amount, record it in table I. 𝑄 ℎ?? 𝑤𝑎??? = ? ℎ??𝑤𝑎??? ∙ ? 𝑤𝑎??? ∙ Δ𝑇 120.92 g x 1 cal x 43.6 °C= 5272.11 Q Determine the Total Heat Lost and the Total heat gained 𝑄 ?𝑎𝑖??? = ?𝑎? 𝑄 ???? = ?𝑎? The heat gain of the cold water and calorimeter cup, theoretically, are equal to the heat loss of the hot water. Compare Heat Gain and Heat Loss finding the percent difference: %?𝑖???????? = |𝑄 ???? − 𝑄 ?𝑎𝑖??? | 1 2 (𝑄 ???? + 𝑄 ?𝑎𝑖??? ) × 100% Was heat lost to or gained from surroundings? Why?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help