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AP Physics B : Thermodynamics and Heat

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Example Questions

Example Question #1 : Thermodynamics And Heat

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$Q=mc\Delta T$

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

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

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 #71 : Ap Physics B

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

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

Possible Answers:

13kg

300kg

13g

300g

12kg

Correct answer:

13kg

Explanation:

Our equation for temperature change using specific heat is:

$Q = mc\Delta t$

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.

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

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

62^oC-54^oC=8^oC

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

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

Use this value back in the original calculation.

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

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

Convert to kilograms to simplify.

$12953.4g\approx13kg$

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Example Question #1 : Thermodynamics And Heat

700 kJ of heat energy is applied to a 2kg block of ice initially at 271 K . This is enough to change the ice completely to water. What is the final temperature of the water after the 700 kJ of heat is applied?

$c_{ice}=2.050\frac{kJ}{kg\cdot K}$

$c_{water}=4.816\frac{kJ}{kg\cdot K}$

$\Delta H^o_{f\ H_2O}=334\frac{kJ}{kg}$

Possible Answers:

275.47K

287.71 K

278.80 K

284.62 K

274.82 K

Correct answer:

275.47K

Explanation:

Since the ice melts, this action occurs in three stages.

1) The heating of the ice to its melting point

2) The melting of the ice into water

3) The heating of the water after it is melted

When a temperature change is involved, the relationship of heat input to temperature change is given by:

$Q = mc \Delta T$

In this formula, is the specific heat (capacity) for the substance. For the phase change involved in melting, there is no temperature change (the energy goes into breaking bonds), and the amount of heat required to completely melt the substance is:

$Q=m\Delta H^o_f$

So this situation is encompassed by three expressions, representing the three parts of the problem, each of which deals with the same mass of material, . Keep in mind that the melting point for ice is 0^oC , or 273K .

1) $Q_{ice}=mc_{ice}\Delta T_{ice}$

$Q_{ice}=(2kg)(2.050\frac{kJ}{kg\cdot K})(273K-271K)$

$Q_{ice}=8.2kJ$

2) $Q_{melt}=m\Delta H^o_{f\ H_2O}$ (for the ice completely melting)

$Q_{melt}=(2kg)(334\frac{kJ}{kg})$

$Q_{melt}=668kJ$

3) $Q_{water}=mc_{water} \Delta T_{water}$ (for the water warming up to its final temperature)

$Q_{water}=(2kg)(4.816\frac{kJ}{kg\cdot K})(T_f-273K)$

The values for $c_{ice}$ , $L_{water}$ , and $c_{water}$ are given, and it is known that $Q_{ice} + Q_{melt} + Q_{water} = 700 kJ$ .

$8.2kJ+668kJ+(2kg)(4.816\frac{kJ}{kg\cdot K})(T_f-273K)=700kJ$

Using this equation we can solve for the final temperature.

$676.2kJ+(9.632\frac{kJ}{K})(T_f-273K)=700kJ$

$T_f-273K=\frac{23.8kJ}{9.632\frac{kJ}{K}}=2.47K$

T_f=275.47K

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AP Physics B : Thermodynamics and Heat

Study concepts, example questions & explanations for AP Physics B

All AP Physics B Resources

Example Questions

Example Question #1 : Thermodynamics And Heat

Example Question #71 : Ap Physics B

Example Question #1 : Thermodynamics And Heat

All AP Physics B Resources

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