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MCAT Physical : Units and Unit Conversion

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Example Questions

Example Question #51 : Stoichiometry And Analytical Chemistry

Which of the following equations is equal to Young's Modulus for a solid?

Possible Answers:

$\frac{A\Delta L}{FL}$

$\frac{F\Delta L}{AL}$

$\frac{FA}{L\Delta L}$

$\frac{FL}{A\Delta L}$

Correct answer:

$\frac{FL}{A\Delta L}$

Explanation:

Young's Modulus is equal to the stress over strain on an object. The stress on an object refers to the pressure that is put on an object, and can be given the below equation.

$Stress=\frac{F}{A}$

The strain on an object refers to the amount an object is deformed in the direction of the force. It can be given the below equation, where $\Delta L$ is the change in length and is the original length.

$Strain=\frac{\Delta L}{L}$

Combining these two equations we can determine Young's Modulus.

$\frac{stress}{strain}= \frac{\frac{F}{A}}{\frac{\Delta L}{L}}= \frac{FL}{A\Delta L}$

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Example Question #52 : Stoichiometry And Analytical Chemistry

Which answer choice is the highest temperature?

Possible Answers:

100K

42^oC

140^oC

400K

150^oF

Correct answer:

140^oC

Explanation:

To find the right answer, we need to know two formulas:

K = C + 273

F = (1.8)(^oC) +32

The simplest way to compare these choices will be to convert them all to Kelvin.

100K

400K

42^oC = 42+ 273 = 315K

150^oF = (1.8)(^oC) + 32

$\rightarrow ^oC=\frac{150-32}{1.8}=65.6^oC$

$\rightarrow 65.6+273=338.6K$

140^oC = 140 + 273 = 413K

This makes our answer 140^oC .

100K<42^oC<150^oF<400K<140^oC

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Example Question #53 : Stoichiometry And Analytical Chemistry

The freezing point of water is $0^{\circ}C$ and the boiling point of water is $100^{\circ}C$ . What is the difference between the boiling point and melting point of water if a student measured the temperatures in Fahrenheit?

$^oF=(\frac{9}{5})^oC+32$

Possible Answers:

$60^{\circ}F$

$100^{\circ}F$

$180^{\circ}F$

$220^{\circ}F$

Correct answer:

$180^{\circ}F$

Explanation:

In this question, you will have to use the given equation twice; once for melting point and once for boiling point. Let’s first convert the melting point to Fahrenheit. The question states that the melting point of water is $0^{\circ}C$ . Plugging this into the equation gives:

$^oF=(\frac{9}{5})^oC+32$

$^oF=(\frac{9}{5})(0^oC)+32$

$F = 32^{\circ}F$

The melting point of water is $32^{\circ}F$ . Similarly, we convert the boiling point of water to Fahrenheit:
$^oF=(\frac{9}{5})(100^oC)+32$

^oF=180+32

$F = 212^{\circ}F$

The boiling point is $212^{\circ}F$ . The difference between boiling point and melting point in Fahrenheit is:
$212^{\circ}F - 32^{\circ}F = 180^{\circ}F$

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Example Question #54 : Stoichiometry And Analytical Chemistry

Which of the following is true regarding the definition of Kelvin?

Possible Answers:

It is defined so that the temperature at which water exists in all three phases (solid, liquid, and gas) is absolute zero

It is defined as the ratio of the melting point to the boiling point of water

It is defined so that the temperature at which water exists in all three phases (solid, liquid, and gas) is 273.16K

It is defined as the ratio of the boiling point to the melting point of water

Correct answer:

It is defined so that the temperature at which water exists in all three phases (solid, liquid, and gas) is 273.16K

Explanation:

Kelvin is a temperature scale that is named after Lord Kelvin. Lord Kelvin, through his experiments, found that absolute zero (0K) occurs at a temperature of $-273.15^{\circ}C$ . Later, scientists defined Kelvin as the fraction $(\frac{1}{273.16})$ of the temperature of water at its triple point (the point on a phase diagram at which water exists as a solid, a liquid, and a gas). This is because the temperature at the triple point of water, in Celsius, is $0.01^{\circ}C$ . In Kelvin scale, the triple point of water occurs at 273.16K .

273.15+0.01^oC=273.16K

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Example Question #55 : Stoichiometry And Analytical Chemistry

The melting point of iron is $2800^{\circ}F$ . In Celsius the melting point of iron is __________ and in Kelvin it is __________.

$^oF=(\frac{9}{5})^oC+32$

Possible Answers:

$1811^oC\ .\ .\ .\ 1538K$

$2400^oC\ .\ .\ .\ 1433K$

$1433^oC\ .\ .\ .\ 2400K$

$1538^oC\ .\ .\ .\ 1811K$

Correct answer:

$1538^oC\ .\ .\ .\ 1811K$

Explanation:

The question states that the melting point of iron is $2800^{\circ}F$ . We can use the given equation to convert this to Celsius.

$^oF=(\frac{9}{5})^oC+32$

$^oC=(^oF-32)(\frac{5}{9})$

$^oC=(2800^oF-32)(\frac{5}{9})$

C=1538^oC

This means that the melting point of iron is $1538^{\circ}C$ . At this point you can choose the right answer because there is only one answer option that has $1538^{\circ}C$ ; however, let’s solve for Kelvin. The second equation you need to know is the conversion from Celsius to Kelvin:

K=^oC+273.15

The melting point of iron in Kelvin is:

K=1538^oC+273.15=1811K

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MCAT Physical : Units and Unit Conversion

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #51 : Stoichiometry And Analytical Chemistry

Example Question #52 : Stoichiometry And Analytical Chemistry

Example Question #53 : Stoichiometry And Analytical Chemistry

Example Question #54 : Stoichiometry And Analytical Chemistry

Example Question #55 : Stoichiometry And Analytical Chemistry

All MCAT Physical Resources

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