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An aluminium block has a volume of 0,420 m³ at a temperature of 298 K. If its temperature rises to 358 K, calculate the increase in volume of the aluminium block.

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4 parts
8.1

0.00174 m3\text{0.00174 m}^3

8.2

5.13 kg\text{5.13 kg}

8.3

4.65 kg\text{4.65 kg}

8.4

20.12 ml\text{20.12 ml}

Step-by-step solution

15 steps

Part 8

HEAT

Part 8.1

0.00174 m3\text{0.00174 m}^3

  • Given: Initial volume V0=0.420 m3V_0 = 0.420 \text{ m}^3. Initial temperature T0=298 KT_0 = 298 \text{ K}. Final temperature Tf=358 KT_f = 358 \text{ K}.
  • Assumption: The volumetric thermal expansion coefficient for aluminium is not provided. We will use a standard value for the linear expansion coefficient of aluminium, αAl=23×10−6 K−1\alpha_{Al} = 23 \times 10^{-6} \text{ K}^{-1}. The volumetric expansion coefficient is then γAl=3αAl\gamma_{Al} = 3\alpha_{Al}.

  1. Calculate the change in temperature.

    ΔT=Tf−T0=358 K−298 K=60 K\Delta T = T_f - T_0 = 358 \text{ K} - 298 \text{ K} = 60 \text{ K}

  2. Calculate the volumetric expansion coefficient for aluminium.

    γAl=3×αAl=3×(23×10−6 K−1)=69×10−6 K−1\gamma_{Al} = 3 \times \alpha_{Al} = 3 \times (23 \times 10^{-6} \text{ K}^{-1}) = 69 \times 10^{-6} \text{ K}^{-1}

  3. Calculate the increase in volume.

    ΔV=γAlV0ΔT\Delta V = \gamma_{Al} V_0 \Delta T ΔV=(69×10−6 K−1)×(0.420 m3)×(60 K)\Delta V = (69 \times 10^{-6} \text{ K}^{-1}) \times (0.420 \text{ m}^3) \times (60 \text{ K}) ΔV=0.0017388 m3\Delta V = 0.0017388 \text{ m}^3

    The increase in volume of the aluminium block is 0.00174 m3\boxed{\text{0.00174 m}^3}.

Part 8.2

5.13 kg\text{5.13 kg}

  • Given: Volume V=0.85 m3V = 0.85 \text{ m}^3. Temperature T=27∘CT = 27^\circ \text{C}. Pressure P=520 kPaP = 520 \text{ kPa}. Gas constant R=287 J/kg⋅KR = 287 \text{ J/kg}\cdot\text{K}.

  1. Convert temperature to Kelvin and pressure to Pascals.

    T=27∘C+273=300 KT = 27^\circ \text{C} + 273 = 300 \text{ K} P=520 kPa=520×103 PaP = 520 \text{ kPa} = 520 \times 10^3 \text{ Pa}

  2. Use the ideal gas law PV=mRTPV = mRT to calculate the mass of the gas.

    m=PVRTm = \frac{PV}{RT} m=(520×103 Pa)×(0.85 m3)(287 J/kg⋅K)×(300 K)m = \frac{(520 \times 10^3 \text{ Pa}) \times (0.85 \text{ m}^3)}{(287 \text{ J/kg}\cdot\text{K}) \times (300 \text{ K})} m=44200086100≈5.1336 kgm = \frac{442000}{86100} \approx 5.1336 \text{ kg}

    The mass of the gas is 5.13 kg\boxed{\text{5.13 kg}}.

Part 8.3

4.65 kg\text{4.65 kg}

  • Given: Initial mass m1=15 kgm_1 = 15 \text{ kg} of CO2_2. Initial pressure P1=150 kPaP_1 = 150 \text{ kPa}. Initial temperature T1=18∘CT_1 = 18^\circ \text{C}. Final pressure P2=210 kPaP_2 = 210 \text{ kPa}. Final temperature T2=38∘CT_2 = 38^\circ \text{C}.
  • Assumption: The specific gas constant for CO2_2 is not given. We will use the universal gas constant Ru=8.314 J/mol⋅KR_u = 8.314 \text{ J/mol}\cdot\text{K} and the molar mass of CO2_2 (MCO2=44.01 g/mol=0.04401 kg/molM_{CO_2} = 44.01 \text{ g/mol} = 0.04401 \text{ kg/mol}) to find the specific gas constant RCO2=Ru/MCO2≈188.90 J/kg⋅KR_{CO_2} = R_u/M_{CO_2} \approx 188.90 \text{ J/kg}\cdot\text{K}. However, since the volume of the cylinder is constant and the gas constant for CO2_2 is the same for both states, it will cancel out in the calculation.

  1. Convert temperatures to Kelvin and pressures to Pascals.

    T1=18∘C+273=291 KT_1 = 18^\circ \text{C} + 273 = 291 \text{ K} P1=150 kPa=150×103 PaP_1 = 150 \text{ kPa} = 150 \times 10^3 \text{ Pa} T2=38∘C+273=311 KT_2 = 38^\circ \text{C} + 273 = 311 \text{ K} P2=210 kPa=210×103 PaP_2 = 210 \text{ kPa} = 210 \times 10^3 \text{ Pa}

  2. Use the ideal gas law for both states.

    Since the volume VV is constant:

    V=m1RT1P1=m2RT2P2V = \frac{m_1 R T_1}{P_1} = \frac{m_2 R T_2}{P_2}

    The gas constant RR cancels out:

    m1T1P1=m2T2P2\frac{m_1 T_1}{P_1} = \frac{m_2 T_2}{P_2}

  3. Solve for the final mass m2m_2.

    m2=m1(P2P1)(T1T2)m_2 = m_1 \left(\frac{P_2}{P_1}\right) \left(\frac{T_1}{T_2}\right) m2=15 kg×(210×103 Pa150×103 Pa)×(291 K311 K)m_2 = 15 \text{ kg} \times \left(\frac{210 \times 10^3 \text{ Pa}}{150 \times 10^3 \text{ Pa}}\right) \times \left(\frac{291 \text{ K}}{311 \text{ K}}\right) m2=15×1.4×0.935691≈19.6495 kgm_2 = 15 \times 1.4 \times 0.935691 \approx 19.6495 \text{ kg}

  4. Calculate the mass of carbon dioxide that was added.

    Δm=m2−m1=19.6495 kg−15 kg=4.6495 kg\Delta m = m_2 - m_1 = 19.6495 \text{ kg} - 15 \text{ kg} = 4.6495 \text{ kg}

    The mass of carbon dioxide that was added is 4.65 kg\boxed{\text{4.65 kg}}.

Part 8.4

20.12 ml\text{20.12 ml}

  • Given: Initial container capacity Vc0=600 ml=600×10−6 m3V_{c0} = 600 \text{ ml} = 600 \times 10^{-6} \text{ m}^3. Initial temperature T0=18∘CT_0 = 18^\circ \text{C}. Final temperature Tf=48∘CT_f = 48^\circ \text{C}. Linear expansion coefficient of container αc=10.8×10−6 K−1\alpha_c = 10.8 \times 10^{-6} \text{ K}^{-1}. Volumetric expansion coefficient of toluene γt=1.15×10−3 K−1\gamma_t = 1.15 \times 10^{-3} \text{ K}^{-1}.

  1. Calculate the change in temperature.

    ΔT=Tf−T0=48∘C−18∘C=30∘C=30 K\Delta T = T_f - T_0 = 48^\circ \text{C} - 18^\circ \text{C} = 30^\circ \text{C} = 30 \text{ K}

  2. Calculate the volumetric expansion coefficient of the container.

    γc=3αc=3×(10.8×10−6 K−1)=32.4×10−6 K−1\gamma_c = 3\alpha_c = 3 \times (10.8 \times 10^{-6} \text{ K}^{-1}) = 32.4 \times 10^{-6} \text{ K}^{-1}

  3. Calculate the increase in volume of the toluene.

    Since the container is completely filled, the initial volume of toluene Vt0V_{t0} is equal to the initial volume of the container Vc0V_{c0}.

    ΔVt=γtVt0ΔT=(1.15×10−3 K−1)×(600×10−6 m3)×(30 K)\Delta V_t = \gamma_t V_{t0} \Delta T = (1.15 \times 10^{-3} \text{ K}^{-1}) \times (600 \times 10^{-6} \text{ m}^3) \times (30 \text{ K}) ΔVt=2.07×10−5 m3\Delta V_t = 2.07 \times 10^{-5} \text{ m}^3

  4. Calculate the increase in volume of the container.

    ΔVc=γcVc0ΔT=(32.4×10−6 K−1)×(600×10−6 m3)×(30 K)\Delta V_c = \gamma_c V_{c0} \Delta T = (32.4 \times 10^{-6} \text{ K}^{-1}) \times (600 \times 10^{-6} \text{ m}^3) \times (30 \text{ K}) ΔVc=5.832×10−7 m3\Delta V_c = 5.832 \times 10^{-7} \text{ m}^3

  5. Calculate the volume of toluene that will overflow.

    ΔVoverflow=ΔVt−ΔVc\Delta V_{overflow} = \Delta V_t - \Delta V_c ΔVoverflow=2.07×10−5 m3−5.832×10−7 m3\Delta V_{overflow} = 2.07 \times 10^{-5} \text{ m}^3 - 5.832 \times 10^{-7} \text{ m}^3 ΔVoverflow=(20.7×10−6 m3)−(0.5832×10−6 m3)\Delta V_{overflow} = (20.7 \times 10^{-6} \text{ m}^3) - (0.5832 \times 10^{-6} \text{ m}^3) ΔVoverflow=20.1168×10−6 m3\Delta V_{overflow} = 20.1168 \times 10^{-6} \text{ m}^3

  6. Convert the overflow volume to millilitres.

    ΔVoverflow=20.1168×10−6 m3×1000 L1 m3×1000 ml1 L=20.1168 ml\Delta V_{overflow} = 20.1168 \times 10^{-6} \text{ m}^3 \times \frac{1000 \text{ L}}{1 \text{ m}^3} \times \frac{1000 \text{ ml}}{1 \text{ L}} = 20.1168 \text{ ml}

    The volume of toluene that will overflow is 20.12 ml\boxed{\text{20.12 ml}}.

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