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.

QUESTION 8: HEAT

8.1

• Given: Initial volume $V_0 = 0.420 m^3$. Initial temperature $T_0 = 298 \text{ K}$. Final temperature $T_f = 358 \text{ K}$.
• Assumption: The volumetric thermal expansion coefficient for aluminium is not provided. We will use a standard value for the linear expansion coefficient of aluminium, $\alpha_{Al} = 23 \times 10^{-6} K^{-1}$. The volumetric expansion coefficient is then $\gamma_{Al} = 3\alpha_{Al}$.

Step 1: Calculate the change in temperature. $\Delta T = T_f - T_0 = 358 K - 298 K = 60 K$ Step 2: Calculate the volumetric expansion coefficient for aluminium. $\gamma_{Al} = 3 \times \alpha_{Al} = 3 \times (23 \times 10^{-6} K^{-1}) = 69 \times 10^{-6} K^{-1}$ Step 3: Calculate the increase in volume. $\Delta V = \gamma_{Al} V_0 \Delta T$ $\Delta V = (69 \times 10^{-6} K^{-1}) \times (0.420 m^3) \times (60 K)$ $\Delta V = 0.0017388 m^3$

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

8.2

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

Step 1: Convert temperature to Kelvin and pressure to Pascals. $T = 27^\circ C + 273 = 300 K$ $P = 520 kPa = 520 \times 10^3 Pa$ Step 2: Use the ideal gas law $PV = mRT$ to calculate the mass of the gas. $m = \frac{PV}{RT}$ $m = \frac{(520 \times 10^3 Pa) \times (0.85 m^3)}{(287 J/kg\cdotK) \times (300 K)}$ $m = \frac{442000}{86100} \approx 5.1336 kg$

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

8.3

• Given: Initial mass $m_1 = 15 \text{ kg}$ of CO$_2$. Initial pressure $P_1 = 150 \text{ kPa}$. Initial temperature $T_1 = 18^\circ \text{C}$. Final pressure $P_2 = 210 \text{ kPa}$. Final temperature $T_2 = 38^\circ \text{C}$.
• Assumption: The specific gas constant for CO$_2$ is not given. We will use the universal gas constant $R_u = 8.314 J/mol\cdot\text{K}$ and the molar mass of CO$_2$ ($M_{CO_2} = 44.01 g/mol = 0.04401 \text{ kg/mol}$) to find the specific gas constant $R_{CO_2} = R_u/M_{CO_2} \approx 188.90 J/kg\cdot\text{K}$. However, since the volume of the cylinder is constant and the gas constant for CO$_2$ is the same for both states, it will cancel out in the calculation.

Step 1: Convert temperatures to Kelvin and pressures to Pascals. $T_1 = 18^\circ C + 273 = 291 K$ $P_1 = 150 kPa = 150 \times 10^3 Pa$ $T_2 = 38^\circ C + 273 = 311 K$ $P_2 = 210 kPa = 210 \times 10^3 Pa$ Step 2: Use the ideal gas law for both states. Since the volume $V$ is constant: $V = \frac{m_1 R T_1}{P_1} = \frac{m_2 R T_2}{P_2}$ The gas constant $R$ cancels out: $\frac{m_1 T_1}{P_1} = \frac{m_2 T_2}{P_2}$ Step 3: Solve for the final mass $m_2$. $m_2 = m_1 \left(\frac{P_2}{P_1}\right) \left(\frac{T_1}{T_2}\right)$ $m_2 = 15 kg \times \left(\frac{210 \times 10^3 Pa}{150 \times 10^3 Pa}\right) \times \left(\frac{291 K}{311 K}\right)$ $m_2 = 15 \times 1.4 \times 0.935691 \approx 19.6495 kg$ Step 4: Calculate the mass of carbon dioxide that was added. $\Delta m = m_2 - m_1 = 19.6495 kg - 15 kg = 4.6495 kg$

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

8.4

• Given: Initial container capacity $V_{c0} = 600 ml = 600 \times 10^{-6} m^3$. Initial temperature $T_0 = 18^\circ \text{C}$. Final temperature $T_f = 48^\circ \text{C}$. Linear expansion coefficient of container $\alpha_c = 10.8 \times 10^{-6} K^{-1}$. Volumetric expansion coefficient of toluene $\gamma_t = 1.15 \times 10^{-3} K^{-1}$.

Step 1: Calculate the change in temperature. $\Delta T = T_f - T_0 = 48^\circ C - 18^\circ C = 30^\circ C = 30 K$ Step 2: Calculate the volumetric expansion coefficient of the container. $\gamma_c = 3\alpha_c = 3 \times (10.8 \times 10^{-6} K^{-1}) = 32.4 \times 10^{-6} K^{-1}$ Step 3: Calculate the increase in volume of the toluene. Since the container is completely filled, the initial volume of toluene $V_{t0}$ is equal to the initial volume of the container $V_{c0}$. $\Delta V_t = \gamma_t V_{t0} \Delta T = (1.15 \times 10^{-3} K^{-1}) \times (600 \times 10^{-6} m^3) \times (30 K)$ $\Delta V_t = 2.07 \times 10^{-5} m^3$ Step 4: Calculate the increase in volume of the container. $\Delta V_c = \gamma_c V_{c0} \Delta T = (32.4 \times 10^{-6} K^{-1}) \times (600 \times 10^{-6} m^3) \times (30 K)$ $\Delta V_c = 5.832 \times 10^{-7} m^3$ Step 5: Calculate the volume of toluene that will overflow. $\Delta V_{overflow} = \Delta V_t - \Delta V_c$ $\Delta V_{overflow} = 2.07 \times 10^{-5} m^3 - 5.832 \times 10^{-7} m^3$ $\Delta V_{overflow} = (20.7 \times 10^{-6} m^3) - (0.5832 \times 10^{-6} m^3)$ $\Delta V_{overflow} = 20.1168 \times 10^{-6} m^3$ Step 6: Convert the overflow volume to millilitres. $\Delta V_{overflow} = 20.1168 \times 10^{-6} m^3 \times \frac{1000 L}{1 m^3} \times \frac{1000 ml}{1 L} = 20.1168 ml$

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

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