AP Physics 2 Flashcards: Specific Heat And Thermal Conductivity

Study Specific Heat And Thermal Conductivity in AP Physics 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Physics 2

Specific Heat And Thermal Conductivity

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QUESTION
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What is the specific heat if q=200 Jq=200 \text{ J}, m=4 kgm=4 \text{ kg}, ΔT=2C\Delta T=2^{\circ}C?

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ANSWER

25 J/kg\cdotpK25 \text{ J/kg·K}. c=q/(mΔT)=200/(4×2)=25c = q/(m\Delta T) = 200/(4 \times 2) = 25

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Flashcard 1: What is the specific heat if q=200 Jq=200 \text{ J}, m=4 kgm=4 \text{ kg}, ΔT=2C\Delta T=2^{\circ}C?

Answer: 25 J/kg\cdotpK25 \text{ J/kg·K}. c=q/(mΔT)=200/(4×2)=25c = q/(m\Delta T) = 200/(4 \times 2) = 25

Flashcard 2: State Fourier's Law of Heat Conduction.

Answer: Heat transfer rate q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x}. Fundamental equation for conductive heat transfer.

Flashcard 3: What does the negative sign in q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x} indicate?

Answer: Heat flows from high to low temperature. Heat flows down temperature gradient naturally.

Flashcard 4: What is the specific heat of water?

Answer: 4.18 J/g°C4.18 \text{ J/g°C}. High value gives water thermal stability.

Flashcard 5: What does the negative sign in q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x} indicate?

Answer: Heat flows from high to low temperature. Heat flows down temperature gradient naturally.

Flashcard 6: Find the specific heat capacity if q=500 Jq=500 \text{ J}, m=2 kgm=2 \text{ kg}, ΔT=5C\Delta T=5^{\circ}C.

Answer: 50 J/kg\cdotpK50 \text{ J/kg·K}. c=q/(mΔT)=500/(2×5)=50c = q/(m\Delta T) = 500/(2 \times 5) = 50

Flashcard 7: What is the thermal conductivity of copper?

Answer: Approximately 385 W/m\cdotpK385 \text{ W/m·K}. High thermal conductivity metal.

Flashcard 8: What is the role of area AA in the thermal conductivity equation?

Answer: Cross-sectional area through which heat flows. Larger area increases heat transfer proportionally.

Flashcard 9: What is the thermal conductivity of air?

Answer: Approximately 0.024 W/m\cdotpK0.024 \text{ W/m·K}. Air is good thermal insulator.

Flashcard 10: State Fourier's Law of Heat Conduction.

Answer: Heat transfer rate q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x}. Fundamental equation for conductive heat transfer.

Flashcard 11: Which material typically has higher thermal conductivity: metal or wood?

Answer: Metal. Metals conduct heat much better than wood.

Flashcard 12: Define specific heat capacity.

Answer: Amount of heat required to raise 1 kg of substance by 1C1^{\circ}C. Intensive property measuring thermal inertia.

Flashcard 13: Find cc if q=750 Jq=750 \text{ J}, m=3 kgm=3 \text{ kg}, ΔT=10C\Delta T=10^{\circ}C.

Answer: 25 J/kg\cdotpK25 \text{ J/kg·K}. c=q/(mΔT)=750/(3×10)=25c = q/(m\Delta T) = 750/(3 \times 10) = 25

Flashcard 14: What is the specific heat if q=200 Jq=200 \text{ J}, m=4 kgm=4 \text{ kg}, ΔT=2C\Delta T=2^{\circ}C?

Answer: 25 J/kg\cdotpK25 \text{ J/kg·K}. c=q/(mΔT)=200/(4×2)=25c = q/(m\Delta T) = 200/(4 \times 2) = 25

Flashcard 15: State the formula for thermal conductivity.

Answer: q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x}. Fourier's Law relates heat flow to material properties.

Flashcard 16: Define specific heat capacity.

Answer: Amount of heat required to raise 1 kg of substance by 1C1^{\circ}C. Intensive property measuring thermal inertia.

Flashcard 17: What is the thermal conductivity of air?

Answer: Approximately 0.024 W/m\cdotpK0.024 \text{ W/m·K}. Air is good thermal insulator.

Flashcard 18: What is the formula for specific heat capacity?

Answer: c=qmΔTc = \frac{q}{m\Delta T}. Rearranged from q=mcΔTq = mc\Delta T.

Flashcard 19: What is thermal conductivity?

Answer: Measure of a material's ability to conduct heat. Material property for heat conduction efficiency.

Flashcard 20: Which material typically has higher thermal conductivity: metal or wood?

Answer: Metal. Metals conduct heat much better than wood.

Flashcard 21: What is the temperature change if 1000 J1000 \text{ J} is added to 2 kg2 \text{ kg} of aluminum (c=900 J/kg\cdotpKc=900 \text{ J/kg·K})?

Answer: 0.56C0.56^{\circ}C. ΔT=q/(mc)=1000/(2×900)=0.56\Delta T = q/(mc) = 1000/(2 \times 900) = 0.56

Flashcard 22: Find cc if q=750 Jq=750 \text{ J}, m=3 kgm=3 \text{ kg}, ΔT=10C\Delta T=10^{\circ}\text{C}.

Answer: 25 J/kg\cdotpK25 \text{ J/kg·K}. c=q/(mΔT)=750/(3×10)=25c = q/(m\Delta T) = 750/(3 \times 10) = 25

Flashcard 23: Identify the meaning of Δx\Delta x in thermal conductivity.

Answer: Thickness of the material. Greater thickness reduces heat transfer rate.

Flashcard 24: What is the effect of high specific heat on temperature change?

Answer: Smaller temperature change for given heat input. More energy needed for same temperature rise.

Flashcard 25: Calculate ΔT\Delta T if q=500 Jq=500 \text{ J}, m=1 kgm=1 \text{ kg}, c=250 J/kg\cdotpKc=250 \text{ J/kg·K}.

Answer: 2C2^{\circ}C. ΔT=q/(mc)=500/(1×250)=2\Delta T = q/(mc) = 500/(1 \times 250) = 2

Flashcard 26: What is the unit of specific heat capacity in SI?

Answer: Joules per kilogram per Kelvin (J/kg·K). Energy per mass per temperature change.

Flashcard 27: What does kk represent in the thermal conductivity formula?

Answer: Thermal conductivity. Material's intrinsic heat conduction property.

Flashcard 28: Find ΔT\Delta T if q=2500 Jq=2500 \text{ J}, m=5 kgm=5 \text{ kg}, c=500 J/kg\cdotpKc=500 \text{ J/kg·K}.

Answer: 1C1^{\circ}C. ΔT=q/(mc)=2500/(5×500)=1\Delta T = q/(mc) = 2500/(5 \times 500) = 1

Flashcard 29: What is the temperature change if 1000 J1000 \text{ J} is added to 2 kg2 \text{ kg} of aluminum (c=900 J/kg\cdotpKc=900 \text{ J/kg·K})?

Answer: 0.56C0.56^{\circ}C. ΔT=q/(mc)=1000/(2×900)=0.56\Delta T = q/(mc) = 1000/(2 \times 900) = 0.56

Flashcard 30: What is the thermal conductivity of copper?

Answer: Approximately 385 W/m\cdotpK385 \text{ W/m·K}. High thermal conductivity metal.

Flashcard 31: What is the unit of specific heat capacity in SI?

Answer: Joules per kilogram per Kelvin (J/kg·K). Energy per mass per temperature change.

Flashcard 32: State the formula for calculating heat transfer.

Answer: q=mcΔTq = mc\Delta T. Relates heat to mass, specific heat, and temperature change.

Flashcard 33: What is the effect of high specific heat on temperature change?

Answer: Smaller temperature change for given heat input. More energy needed for same temperature rise.

Flashcard 34: What is the role of area AA in the thermal conductivity equation?

Answer: Cross-sectional area through which heat flows. Larger area increases heat transfer proportionally.

Flashcard 35: What is the specific heat of water?

Answer: 4.18 J/g°C4.18 \text{ J/g°C}. High value gives water thermal stability.

Flashcard 36: What does qq represent in heat equations?

Answer: Heat transfer. Heat energy or heat flow rate.

Flashcard 37: Find the specific heat if q=360 Jq=360 \text{ J}, m=2 kgm=2 \text{ kg}, ΔT=2C\Delta T=2^{\circ}C.

Answer: 90 J/kg\cdotpK90 \text{ J/kg·K}. c=q/(mΔT)=360/(2×2)=90c = q/(m\Delta T) = 360/(2 \times 2) = 90

Flashcard 38: Find the mass if q=1500 Jq=1500 \text{ J}, c=200 J/kg\cdotpKc=200 \text{ J/kg·K}, ΔT=3C\Delta T=3^{\circ}C.

Answer: 2.5 kg2.5 \text{ kg}. m=q/(cΔT)=1500/(200×3)=2.5m = q/(c\Delta T) = 1500/(200 \times 3) = 2.5

Flashcard 39: Find ΔT\Delta T if q=2500 Jq=2500 \text{ J}, m=5 kgm=5 \text{ kg}, c=500 J/kg\cdotpKc=500 \text{ J/kg·K}.

Answer: 1C1^{\circ}C. ΔT=q/(mc)=2500/(5×500)=1\Delta T = q/(mc) = 2500/(5 \times 500) = 1

Flashcard 40: Find the mass if q=1500 Jq=1500 \text{ J}, c=200 J/kg\cdotpKc=200 \text{ J/kg·K}, ΔT=3C\Delta T=3^{\circ}C.

Answer: 2.5 kg2.5 \text{ kg}. m=q/(cΔT)=1500/(200×3)=2.5m = q/(c\Delta T) = 1500/(200 \times 3) = 2.5

Flashcard 41: Find the specific heat if q=360 Jq=360 \text{ J}, m=2 kgm=2 \text{ kg}, ΔT=2C\Delta T=2^{\circ}C.

Answer: 90 J/kg\cdotpK90 \text{ J/kg·K}. c=q/(mΔT)=360/(2×2)=90c = q/(m\Delta T) = 360/(2 \times 2) = 90

Flashcard 42: What is the unit of thermal conductivity in SI?

Answer: Watts per meter per Kelvin (W/m·K). Power per length per temperature difference.

Flashcard 43: Calculate ΔT\Delta T if q=500 Jq=500 \text{ J}, m=1 kgm=1 \text{ kg}, c=250 J/kg\cdotpKc=250 \text{ J/kg·K}.

Answer: 2C2^{\circ}C. ΔT=q/(mc)=500/(1×250)=2\Delta T = q/(mc) = 500/(1 \times 250) = 2

Flashcard 44: Identify the meaning of Δx\Delta x in thermal conductivity.

Answer: Thickness of the material. Greater thickness reduces heat transfer rate.

Flashcard 45: What is thermal conductivity?

Answer: Measure of a material's ability to conduct heat. Material property for heat conduction efficiency.

Flashcard 46: State the formula for thermal conductivity.

Answer: q=kAΔTΔxq = -kA\frac{\Delta T}{\Delta x}. Fourier's Law relates heat flow to material properties.

Flashcard 47: What does ΔT\Delta T represent in heat transfer formulas?

Answer: Change in temperature. Final minus initial temperature.

Flashcard 48: What is the unit of thermal conductivity in SI?

Answer: Watts per meter per Kelvin (W/m·K). Power per length per temperature difference.

Flashcard 49: What does qq represent in heat equations?

Answer: Heat transfer. Heat energy or heat flow rate.

Flashcard 50: Find the specific heat capacity if q=500 Jq=500 \text{ J}, m=2 kgm=2 \text{ kg}, ΔT=5C\Delta T=5^{\circ}C.

Answer: 50 J/kg\cdotpK50 \text{ J/kg·K}. c=q/(mΔT)=500/(2×5)=50c = q/(m\Delta T) = 500/(2 \times 5) = 50

Flashcard 51: What does kk represent in the thermal conductivity formula?

Answer: Thermal conductivity. Material's intrinsic heat conduction property.

Flashcard 52: Identify the symbol for specific heat capacity.

Answer: cc. Lowercase c is standard notation.

Flashcard 53: What is the formula for specific heat capacity?

Answer: c=qmΔTc = \frac{q}{m\Delta T}. Rearranged from q=mcΔTq = mc\Delta T.

Flashcard 54: What does ΔT\Delta T represent in heat transfer formulas?

Answer: Change in temperature. Final minus initial temperature.

Flashcard 55: State the formula for calculating heat transfer.

Answer: q=mcΔTq = mc\Delta T. Relates heat to mass, specific heat, and temperature change.