Known Data are provided for a constant-volume calorimeter testing cooking oil for calorie value.
Find Determine the change in internal energy of the contents of the calorimeter chamber.
Schematic and Given Data:
Engineering Model
1. The closed system is shown by the dashed line in the accompanying figure.
2. The total volume remains constant, including the chamber, water bath, and the amount of water modeling the metal parts.
3. Water is modeled as incompressible with constant specific heat c.
4. Heat transfer with the surroundings is negligible, and there is no change in kinetic or potential energy.
Analysis With the assumptions listed, the closed system energy balance reads
\Delta U+\Delta K E^{\nearrow0}+\Delta P E^{\nearrow0}=Q^{\nearrow0}-W^{\nearrow0}
or
(\Delta U)_{\text {contents }}+(\Delta U)_{\text {water }}=0
thus
(\Delta U)_{\text {contents }}=-(\Delta U)_{\text {water }} (a)
The change in internal energy of the contents is equal and opposite to the change in internal energy of the water.
Since water is modeled as incompressible, Eq. 3.20a is used to evaluate the right side of Eq. (a), giving
u_{2}-u_{1}=c\left(T_{2}-T_{1}\right) (3.20a)
h_{2}-h_{1}=c\left(T_{2}-T_{1}\right)+\underline{v\left(p_{2}-p_{1}\right)} (incompressible, constant c) (3.20b)
1 2 (\Delta U)_{\text {contents }}=-m_{w} c_{w}\left(T_{2}-T_{1}\right) (b)
With m_{w}=2.15 kg +0.5 kg =2.65 kg ,\left(T_{2}-T_{1}\right)=0.3 K, and c_{w}=4.18 kJ / kg \cdot K from Table A-19, Eq. (b) gives
(\Delta U)_{\text {contents }}=-(2.65 kg )(4.18 kJ / kg \cdot K )(0.3 K )=-3.32 kJ
Converting to kcal, and expressing the result on a per milliliter of oil basis using the oil volume, 0.1 mL, we get
\frac{(\Delta U)_{\text {contents }}}{V_{\text {oil }}}=\frac{-3.32 kJ }{0.1 mL }\left|\frac{1 kcal }{4.1868 kJ }\right|
= -7.9 kcal/mL
The calorie value of the cooking oil is the magnitude—that is, 7.9 kcal/mL. Labels on cooking oil containers usually give calorie value for a serving size of 1 tablespoon (15 mL). Using the calculated value, we get 119 kcal per tablespoon.
1 The change in internal energy for water can be found alternatively using Eq. 3.12 together with saturated liquid internal energy data from Table A-2.
u(T, p) \approx u_{ f }(T) (3.12)
2 The change in internal energy of the chamber contents cannot be evaluated using a specific heat because specific heats are defined (Sec. 3.9) only for pure substances—that is, substances that are unchanging in composition.
Skills Developed
Ability to…
• define a closed system and identify interactions within it and on its boundary.
• apply the energy balance using the incompressible substance model.
Quick Quiz
Using Eq. 3.12 together with saturated liquid internal energy data from Table A-2, find the change in internal energy of the water, in kJ, and compare with the value obtained assuming water is incompressible. Ans. 3.32 kJ
