Chapter 9, Problem 9.37PAE

9.37 A calorimeter contained 75.0 g of water at 16.95°C. A 93.3-g sample of iron at 65.58°C was placed in it, giving a final temperature of 19.68°C for the system. Calculate the heat capacity of the calorimeter. Specific heats are $\text{4}.{\text{184J g}}^{–\text{1}}°{\text{C}}^{–\text{1}}$ for ${\text{H}}_{\text{2}}\text{O}$ and $0.444{\text{J g}}^{–\text{1}}°{\text{C}}^{–\text{1}}$ for Fe.

Interpretation Introduction

Interpretation:

Using the information provided, the solutions can be obtained by calculations using suitable formulas.

Concept Introduction:

• The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. Formula for heat required is-

  $Heatrequired,Q=c\times m\times \Delta T$

  Where

  c = specific heat of the substance

  m = mass of the substance

  $\Delta T$ = Temperature difference

• According to the first law of thermodynamics,

  Q_iron = Q_{calorimetersystem}

  Q_{calorimetersystem}= Q_calorimeter+Q_water

Answer to Problem 9.37PAE

Solution:

Heat capacity of calorimeter = 5.1025 J g-1 ^oC-1

Explanation of Solution

Given information:

Mass of water in the calorimeter = 75.0 g

Temperature of water in the calorimeter = 16.95 ^oC

Mass of sample of iron = 93.3 g

Temperature of sample of iron = 65.58 ^oC

Final temperature = 19.68 ^oC

Specific heat capacity of water = 4.184 J g -1 ^oC-1

Specific heat capacity of iron = 0.444 J g -1 ^oC-1

According to the thermodynamic equation is-

Q_iron = Q_{calorimetersystem}

Q_{calorimetersystem}= Q_calorimeter+Q_water

Where

$Q_{iron}=C_p\times \Delta T\times m$

C_p= Specific heat capacity of iron

$\Delta T$ = Temperature difference

m = mass of the iron

$Q_{calorimetersystem}=C_{p{(}_{calorimetersystem})}\times \Delta T\times m_{(water)}$

$C_p{}_{(water)}$= Specific heat capacity of water

Specific heat of water = 4.184 J g-1 ^oC-1

$\Delta T$ = Temperature difference

m(water) = mass of the titanium water

Therefore

Qiron =Qcalorimeter  systemQcalorimeter system= Qcalorimeter +QwaterCp×ΔT×m=Cp( calorimeter    system)×ΔT×m(water)0.444×(65.58−19.68)×93.3=Cp( calorimeter    system) ×(19.68−16.95)×75.0Cp( calorimeter    system)=0.444×(65.58−19.68)×93.3(19.68−16.95)×75.0Cp( calorimeter    system)=0.444×(45.9)×93.3(2.73)×75.0Cp( calorimeter    system)=1901.4166204.75=9.2865Qcalorimeter system= Qcalorimeter +Qwater9.2865= Qcalorimeter +4.184Qcalorimeter =9.2865−4.184=5.1025

Heat capacity of calorimeter = 5.1025 J g-1 ^oC-1

Conclusion

Heat capacity of calorimeter = 5.1025 J g-1 ^oC-1