GCSE Physics (9-1)
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    8.2 Internal energy, energy transfers and particle motions — Practise

    Work through these on your own. Keep the units with your numbers. Some questions mix a change of temperature with a change of state, so decide which relationship you need for each stage before you calculate.

    Both the questions and the answers below come from a set that is checked by computer every time the book is built.

    1. A tank contains 2.5 kg of water. The water is heated through 40 °C. The specific heat capacity of water is 4200 J/kg°C. Calculate the energy transferred to the water.
    2. A block of metal of mass 1.0 kg gains 23400 J when its temperature rises by 60 °C. Calculate its specific heat capacity.
    3. Calculate the energy needed to melt 1.5 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.3 x 10^5 J/kg.
    4. Calculate the energy needed to boil away 0.30 kg of water at 100 °C. The specific latent heat of vaporisation of water is 2.3 x 10^6 J/kg.
    5. Explain what is meant by the internal energy of a system, and state two things that can happen when the system is heated.
    6. A heating graph for a pure substance has a flat part at 0 °C. Explain what is happening during this flat part, and why the temperature does not change.
    7. Distinguish between specific heat capacity and specific latent heat. Include the unit of each and when each one is used.
    8. A block of mass 1.2 kg is given 50400 J of energy. The specific heat capacity of the block is 4200 J/kg°C. Calculate the temperature rise.
    9. Calculate the total energy needed to warm 0.50 kg of water from 0 °C to 100 °C and then boil it all away. Use c = 4200 J/kg°C and L = 2.3 x 10^6 J/kg.
    10. Compare the energy needed to melt 1.0 kg of ice at 0 °C with the energy needed to boil away 1.0 kg of water at 100 °C. Which is larger, and by roughly how many times?
    11. A heater transfers energy at a rate of 1500 W for 4.0 minutes. The energy is transferred to 2.0 kg of water. Calculate the temperature rise of the water. The specific heat capacity of water is 4200 J/kg°C. This uses power from Chapter 1.
    12. Say which relationships you will use, then calculate the total energy needed to melt 0.25 kg of ice at 0 °C and warm the meltwater to 30 °C. Use L = 3.3 x 10^5 J/kg and c = 4200 J/kg°C.
    Answers and concise solutions

    1. Energy transferred to the water

    Answer: ΔE=mcΔθ=2.5×4200×40=420 000 J=4.2×105 J\Delta E = mc\Delta\theta = 2.5 \times 4200 \times 40 = 420\,000\,\text{J} = 4.2 \times 10^5\,\text{J}.

    2. Specific heat capacity of the metal

    Answer: c=ΔEmΔθ=23 4001.0×60=390 J/kg ∘Cc = \dfrac{\Delta E}{m\Delta\theta} = \dfrac{23\,400}{1.0 \times 60} = 390\,\text{J/kg}\,^\circ\text{C}.

    3. Energy needed to melt the ice

    Answer: E=mL=1.5×3.3×105=495 000 J=5.0×105 JE = mL = 1.5 \times 3.3 \times 10^5 = 495\,000\,\text{J} = 5.0 \times 10^5\,\text{J} to 2 significant figures.

    4. Energy needed to boil the water

    Answer: E=mL=0.30×2.3×106=690 000 J=6.9×105 JE = mL = 0.30 \times 2.3 \times 10^6 = 690\,000\,\text{J} = 6.9 \times 10^5\,\text{J}.

    5. Internal energy and heating

    • Internal energy is the total kinetic energy and the total potential energy of all the particles in the system.
    • Heating can raise the temperature of the system, or it can produce a change of state without changing the temperature.

    6. The flat part of the heating graph

    • The substance is changing state, in this case melting.
    • The energy supplied breaks the forces between the particles, so their potential energy increases.
    • The average kinetic energy of the particles does not change, so the temperature stays constant.

    7. Specific heat capacity and specific latent heat

    • Specific heat capacity is the energy needed to raise the temperature of one kilogram of a substance by one degree Celsius. It is measured in joules per kilogram per degree Celsius, and it is used when the temperature changes.
    • Specific latent heat is the energy needed to change the state of one kilogram of a substance with no change in temperature. It is measured in joules per kilogram, and it is used during a change of state.

    8. Temperature rise of the block

    Answer: Δθ=ΔEmc=50 4001.2×4200=10 ∘C\Delta\theta = \dfrac{\Delta E}{mc} = \dfrac{50\,400}{1.2 \times 4200} = 10\,^\circ\text{C}.

    9. Total energy for the two stages

    Answer: Warming: ΔE=0.50×4200×100=210 000 J\Delta E = 0.50 \times 4200 \times 100 = 210\,000\,\text{J}. Boiling: E=0.50×2.3×106=1 150 000 JE = 0.50 \times 2.3 \times 10^6 = 1\,150\,000\,\text{J}. Total: 1 360 000 J=1.4×106 J1\,360\,000\,\text{J} = 1.4 \times 10^6\,\text{J}.

    10. The two latent heats compared

    • Melting 1.0 kg of ice needs E = mL = 1.0 x 3.3 x 10^5 = 3.3 x 10^5 J.
    • Boiling away 1.0 kg of water needs E = mL = 1.0 x 2.3 x 10^6 = 2.3 x 10^6 J, which is about seven times larger.

    11. Temperature rise of the water

    Answer: Energy =Pt=1500×240=360 000 J= Pt = 1500 \times 240 = 360\,000\,\text{J}. Then Δθ=ΔEmc=360 0002.0×4200=42.9 ∘C=43 ∘C\Delta\theta = \dfrac{\Delta E}{mc} = \dfrac{360\,000}{2.0 \times 4200} = 42.9\,^\circ\text{C} = 43\,^\circ\text{C} to 2 significant figures.

    12. Total energy for melting and warming

    Answer: Melting uses E=mL=0.25×3.3×105=82 500 JE = mL = 0.25 \times 3.3 \times 10^5 = 82\,500\,\text{J}. Warming uses ΔE=mcΔθ=0.25×4200×30=31 500 J\Delta E = mc\Delta\theta = 0.25 \times 4200 \times 30 = 31\,500\,\text{J}. Total: 114 000 J=1.1×105 J114\,000\,\text{J} = 1.1 \times 10^5\,\text{J}.

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