Question

29 NEUTRALİZATION CALORIMETRY 3-THERMODYNAMICS: ENTHALPY OF EXERCISES 1. If the final te or erature taken in part I was higher than the true value, the calculated value for the heat capacity of the lower than the true value? calorimeter be higher than in the process of transferring the metal cylinder from the hot water to the in part II, the cylinder lost some heat, would the calculated value 2. If for the specific heat of the metal be higher or lower than the true valuer 3. A student placed 45.0 g of water into a calorimeter and observed that the tem perature was 20.0°C. A 60.0 g metal cylinder was heated to 92.0°C and then put into the calorimeter. The system was allowed to come to equilibrium. What was the final temperature in the calorimeter if the specific heat of the metal cylinder was 0.150 calKg-C) and the heat capacity of the calorimeter 5.20 cal/°C? was 4. Suppose that a student performed part III B and the calculations perfectl cept for the following. Unknown to the student, the hot water tween the time its temperature was measured and the time it was calorimeter. What effect would this solution? That is, would the calculated value be than the true value? had cooled be- added to the have on the calculated specific heat of the equal to, higher than, or lower 25 3-THERMODYNAMICS: ENTHALPY OF NEUTRALIZATION CALORIMETRY NAME INSTRUCTOR EXPERIMENT 3: REPORT FOR CALORIMETRY DATA/RESULTS . THE HEAT CAPACITY OF THE CALORIMETER *1. mass of calorimeter: (g) *2. mass of calorimeter + cold water: (g) SECTION DATE GRADE example trial 1*rial 2 71.834 73. 048 . /0253 ter + cold water + hot water: (g) 121.621 22.6 500 16260 49.787- 27 4. mass of cold water (ma): 5. mass of hot water (m,H,O):(g) "6 temperature ofcold water (T,HO):(0) *7. temperature of hot water (T): ec) *8. final temperature (T,): (°C) 9. heat capacity of calorimeter (HCa: (cal/C)42 64.5 5 43.5 10. average value of the heat capacity of the c1 calorimeter (HC): (cal/oC)

Answer 1

1. Heat capacity of calorimeter (C) is defined as the amount of heat absorbed by it (q) when temperature of its contents is raised by 1°C.

$C=\frac{q}{\Delta T}$ (${\Delta T}$ is the rise in temperature=T_f​​​​_{​​​inal}​-T_initial)

So if the final temperature taken in part 1 is more than the true value, it will give a larger value of ${\Delta T}$ than the true value. So as per the formula used to obtain C, the value of C will come out to be less than the true value (as $C\alpha \frac{1}{\Delta T}$ ).

So the calculated value of specific heat will be less than the true value.