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Q5.9:
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from (35^{\circ}\text{C}) to (55^{\circ}\text{C}). Molar heat capacity of Al is (24 \text{ J mol}^{-1} \text{ K}^{-1}).

Solution :

Step 1: Given Variables & Conversion to SI Units

To calculate the heat energy required, we identify the following parameters:

  • Mass of Aluminium ($m$): $60.0 \text{ g}$
  • Molar Mass of Aluminium ($M$): $26.98 \text{ g mol}^{-1}$
  • Molar heat capacity ($C_m$): $24 \text{ J mol}^{-1} \text{ K}^{-1}$
  • Initial temperature ($T_i$): $35^{\circ}\text{C} = (35 + 273.15) \text{ K} = 308.15 \text{ K}$
  • Final temperature ($T_f$): $55^{\circ}\text{C} = (55 + 273.15) \text{ K} = 328.15 \text{ K}$

Step 2: Determining the Moles of Aluminium

The number of moles ($n$) is calculated using the ratio of the given mass to the molar mass of the element [Per the definition of the mole]:

$n = \frac{m}{M} = \frac{60.0 \text{ g}}{26.98 \text{ g mol}^{-1}} \approx 2.2239 \text{ mol}$

Step 3: Calculating Temperature Change

The change in temperature ($\Delta T$) is invariant between the Celsius and Kelvin scales [As $\Delta T = (T_f + 273.15) - (T_i + 273.15) = T_f - T_i$]:

$\Delta T = 55^{\circ}\text{C} - 35^{\circ}\text{C} = 20 \text{ K}$

Step 4: Applying the Thermodynamic Heat Equation

The heat energy ($q$) required is defined by the product of the number of moles, the molar heat capacity, and the temperature difference [By the definition of molar heat capacity]:

$q = n \cdot C_m \cdot \Delta T$

$q = (2.2239 \text{ mol}) \times (24 \text{ J mol}^{-1} \text{ K}^{-1}) \times (20 \text{ K})$

$q = 2.2239 \times 480 \text{ J} \approx 1067.47 \text{ J}$

Step 5: Unit Conversion to Kilojoules

To express the final result in kJ, we apply the conversion factor $1 \text{ kJ} = 1000 \text{ J}$:

$q_{\text{kJ}} = \frac{1067.47 \text{ J}}{1000} \approx 1.067 \text{ kJ}$

Energy (J) Temp (K) q = n*Cm*ΔT

Final Solution: The heat required to raise the temperature of 60.0 g of Aluminium is approximately 1.067 kJ.


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