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AP Physics B : Understanding Heat and Temperature

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Example Questions

Example Question #3 : Ap Physics 2

Which of the following is not an acceptable unit for specific heat?

Possible Answers:

$\frac{cal}{g\cdot K}$ $\displaystyle \frac{cal}{g\cdot K}$

$\frac{J}{N\cdot K}$ $\displaystyle \frac{J}{N\cdot K}$

$\frac{Btu}{^oF\cdot lb}$ $\displaystyle \frac{Btu}{^oF\cdot lb}$

$\frac{J}{mol\cdot K}$ $\displaystyle \frac{J}{mol\cdot K}$

$\frac{J}{g\cdot K}$ $\displaystyle \frac{J}{g\cdot K}$

Correct answer:

$\frac{J}{N\cdot K}$ $\displaystyle \frac{J}{N\cdot K}$

Explanation:

Specific heat is most commonly applied in the equation to determine the heat required/released during a temperature change:

$Q=mc\Delta T$ $\displaystyle Q=mc\Delta T$

Rearranging this equal, we can see that this units of specific heat are units of heat per units of mass times units of temperature.

$c=\frac{Q}{m\Delta T}$ $\displaystyle c=\frac{Q}{m\Delta T}$

Heat energy can have the units of Joules or calories. Mass can have units of grams or kilograms. Temperature can have units of Kelvin, degrees Celsius, or degrees Fahrenheit. British thermal units (Btu) is a less common unit of heat, defined by the amount of heat required to raise one pound of water by one degree Fahrenheit.

Newtons are a unit of force, and cannot be used in the units for specific heat.

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Example Question #4 : Ap Physics 2

A sample of copper is heated with 40000J $\displaystyle 40000J$ of heat and the temperature of the sample rises from 54^oC $\displaystyle 54^oC$ to 62^oC $\displaystyle 62^oC$. What mass of copper is present in this sample?

$c_{Cu}=0.386\frac{J}{g\cdot K}$ $\displaystyle c_{Cu}=0.386\frac{J}{g\cdot K}$

Possible Answers:

13g $\displaystyle 13g$

300g $\displaystyle 300g$

13kg $\displaystyle 13kg$

12kg $\displaystyle 12kg$

300kg $\displaystyle 300kg$

Correct answer:

13kg $\displaystyle 13kg$

Explanation:

Our equation for temperature change using specific heat is:

$Q = mc\Delta t$ $\displaystyle Q = mc\Delta t$

Since Celsius and Kelvin use the same interval scale, we do not need to convert to find the change in temperature. Using the initial and final temperatures, the energy input, and the specific heat of copper, we can calculate the mass of copper present.

$40000J=m(0.386\frac{J}{g\cdot K})(62^oC-54^oC)$ $\displaystyle 40000J=m(0.386\frac{J}{g\cdot K})(62^oC-54^oC)$

To simplify, we will solve for the temperature change separately.

$\displaystyle 62^oC-54^oC=8^oC$

Since the scale for Celsius and Kelvin is the same, this change is the same in either scale.

$\Delta 8^oC=\Delta 8K$ $\displaystyle \Delta 8^oC=\Delta 8K$

Use this value back in the original calculation.

$40000J=m(0.386\frac{J}{g\cdot K})(8K)$ $\displaystyle 40000J=m(0.386\frac{J}{g\cdot K})(8K)$

$m=\frac{40000J}{3.088\frac{J}{g}}=12953.4g$ $\displaystyle m=\frac{40000J}{3.088\frac{J}{g}}=12953.4g$

Convert to kilograms to simplify.

$12953.4g\approx13kg$ $\displaystyle 12953.4g\approx13kg$

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AP Physics B : Understanding Heat and Temperature

Study concepts, example questions & explanations for AP Physics B

All AP Physics B Resources

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Example Question #3 : Ap Physics 2

Example Question #4 : Ap Physics 2

All AP Physics B Resources

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