T271 - iCMA 41: Electric Vehicle Battery, Microbiology and Dimensions.

This assignment solution provides detailed answers to questions from T271 iCMA 41. It includes calculations for the energy stored in an electric vehicle battery, determination of bacterial population growth from a log-linear graph, dimensional analysis of stress/strain and kinematic equations, gradient calculation from a log-log plot, energy required to break bonds in carbon dioxide, and analysis of a log-log graph to find gradient and y-intercept. Numerical answers are provided with appropriate units and significant figures. Desklib is the perfect platform to find similar assignments and study resources.

T271 - iCMA 41
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Question 1
An electric vehicle battery pack is rated as having a useable energy capacity of 26.5 kWh. How
much useable energy is stored by the battery at full charge?
Give your answer exactly, in megajoules (MJ).
Energy stored by battery:
Solution
1Kwh = 3.6 MJ, Thus, 26.5kWh 26.5× 3.6
1 =95.4 kWh
Question 2
In a microbiological culture, the population of bacteria proved to increase exponentially with
time as shown in the following log-linear graph.
The plot shows the number N of bacteria in a culture at time t seconds. Interpreting this type of
graph is important to your studies in T271.
Find
the

number of bacteria at time 2 s. Convert your answer to give the number of bacteria as a power of
10, stating the exponent to 2 significant figures.
(You should read the plot as accurately as you can, but a tolerance of ± 1 will be allowed in the
second significant figure in your answer).
Solution
1.9×102=number of bacteria at 2 seconds
Number of bacteria at 2 sec =19
Number of bacteria N = 10 ͯ where x =
N=19; x=log(19)=1.2788 to 2 significant figures: x= 1.3
Question 3
What are the dimensions of the following two expressions?
Please note that both parts of the question need to be answered or a score of zero will be awarded
for the question.
Provided both parts of the question are attempted, if only one part is correct, half-marks will be
awarded.
(a) stress/strain = σ /ε
 (M) (L ̄ ¹) (T ̄ ²)
 (L) (T ̄ ²)
 (M) (L) (T ̄ ²)
 (L ̄ ¹)
Solution
Strain is change in configuration, thus it is dimensionless. Therefore, k has dimension of stress.
Stress = force
area The dimension of stress is (M) (L ̄ ¹) (T ̄ ²). Thus, dimension of stress/strain =
σ /ε is (M) (L ̄ ¹) (T ̄ ²).
(b) Velocity x time + ½ x acceleration x time² = ut + ½ at ²
 (L) (T)
 (L)
 (L) (T ̄ ¹)
 (L) (T ̄ ²)
Solution
Dimension of Size=L and Velocity=L\T( Wilkins & Pennel, 1995).

The numbers have no dimensions e.g. 1
2 ¿ M0 L0 T 0=1. so, ut + ½ at ² and velocity x time
+ ½ x acceleration x time² both are equal to L1=L thus it’s dimension is (L)
Question 4
The following graph is plotted on log-log axes and such graphs are used extensively throughout
Part 1 of the module in the material selection charts.
The line on the graph above passes through the points (79.43, 0.1072) and (1259, 0.7413).
Find the gradient of the line, giving your answer to 1 decimal place.
Gradient:
Solution
Slope of a log-log plot
log10 F ( x ) =m log10 x +b, F(x) = xm .10b To find the gradient, two points are selected on x-axis
say x1∧x2 using the above equation.

log f ¿ ¿) = mlog ( x1 ) +b andlog f ¿ ¿) = mlog ( x2 ) +b .The gradient m;
log ( y2 ) −¿ log ( y1)
log ( x2 ) −log (x1) ¿= m=
log y2
y1
log x2
x1
. x1=79.43 x2=1259 y1=0.1072 y2=0.7413
Log( 1259
79.43 )=1.2000 Log ( 0.7413
0.1072 )=0.8398 m= 0.8398
1.2000 =0.6998
To 1 decimal point= 0.7
Question 5
Carbon dioxide (CO2) is a covalently bonded molecule, made up of two C=O bonds.
Calculate the energy required to break all of the bonds in 148 grammes (g) of (CO2).
Use the following values:
 C=O bond energy = 741 kJ mol̄ ¹
 Atomic mass of carbon = 12.0115 g mol̄ ¹
 Atomic mass of oxygen = 15.999 g mol̄ ¹
Give your answer in kilojoules (kJ) to three significant figures.
Energy required: kJ
Solution
1 bond of C=O require 741kJ mol̄ ¹ but will be forming 2 moles of C=O bonds (Ley &
Thomas, 2003) which will require 2 ×741=1482kJ.
1 mole of carbon dioxide= (2×15.999+12.0115 ¿=44.0095 g
1 mole of CO2 =44.0095g, so how many moles are in 148g of CO2; 148 g
44.0095 g =3.3603
moles
1 mole require 1482kJ to break what about 3.2720 moles; 3.3603 moles ×1482 kJ
1 mole
=4979.9646kJ= 4980 kJ (3 significant figures).

Question 6
The log-log graph below shows how a certain quantity A varies with another quantity x.
For the
line
shown,
find its
gradient
and the
value of
A at the
y-
intercept.
Give your answers to 2 significant figures. You may ignore the units of the quantities involved.
(You should read the plot as accurately as you can, but a tolerance of ± 1 will be allowed in the
second significant digit in your answer).
Please note that both parts of the question need to be answered or a score of zero will be awarded
for the question.
Provided both parts of the question are attempted, if only one part is correct, half-marks will be
awarded.

Gradient:
Solution
log y= k log x +¿ log a ¿ . Y= mX+b where m=k is the gradient and b=log a is the y intercept.
m=log ( y2 )−¿ log ( y1)
log ( x2 )−log ( x1) ¿=( 0.6−0.1
0.3−0.0 )=1.6667 to 2 significant figures = 1.7
Value of A at intercept
Solution
A=log a
log y= k log x +¿ log a ¿
At log x=0.3, log y=0.6
From log y= k log x +¿ log a ¿ we have 0.6 = 1.7(0.3) +log a
Log a= 0.6-0.51= 0.09
Thus, A at intercept = 0.1 in 2 significant figures

References
Ley, S. V., & Thomas, A. W. (2003). Modern Synthetic Methods for Copper‐Mediated
Bond Formation. Angewandte Chemie International Edition, 42(44), 5400-5449
Wilkins, M. D., Abriola, L. M., & Pennell, K. D. (1995). An experimental investigation of rate
‐
limited nonaqueous phase liquid volatilization in unsaturated porous media: Steady state mass
transfer. Water Resources Research, 31(9), 2159-2172.