Calculate the number of `kJ` of heat necessary to raise the temperature of `60.0 g` of aluminium from `35^(@)C` to `55^(@)C`. Molar heat capacity of `Al` is `24 J g m^(-1)`.

Explanation

To calculate the number of kilojoules of heat necessary to raise the temperature of 60.0 g of aluminum from 35°C to 55°C, we can follow these steps:

### Step 1: Identify the given values
- Mass of aluminum (m) = 60.0 g
- Initial temperature (T1) = 35°C
- Final temperature (T2) = 55°C
- Molar heat capacity of aluminum (C) = 24 J/g·°C

### Step 2: Calculate the change in temperature (ΔT)
\[ \Delta T = T2 - T1 = 55°C - 35°C = 20°C \]

### Step 3: Calculate the number of moles (n) of aluminum
To find the number of moles, we need the molar mass of aluminum, which is approximately 27 g/mol.
\[ n = \frac{m}{\text{Molar mass}} = \frac{60.0 \, \text{g}}{27 \, \text{g/mol}} \approx 2.22 \, \text{mol} \]

### Step 4: Calculate the heat (Q) required using the formula
The formula for heat is:
\[ Q = n \cdot C \cdot \Delta T \]
Substituting the values:
\[ Q = 2.22 \, \text{mol} \cdot 24 \, \text{J/g·°C} \cdot 20 \, \text{°C} \]

### Step 5: Calculate Q in Joules
\[ Q = 2.22 \cdot 24 \cdot 20 = 1065.6 \, \text{J} \]

### Step 6: Convert Joules to kilojoules
Since 1 kJ = 1000 J, we convert Joules to kilojoules:
\[ Q = \frac{1065.6 \, \text{J}}{1000} = 1.0656 \, \text{kJ} \]

### Step 7: Round the answer
Rounding to two decimal places, we get:
\[ Q \approx 1.07 \, \text{kJ} \]

### Final Answer
The heat necessary to raise the temperature of 60.0 g of aluminum from 35°C to 55°C is approximately **1.07 kJ**.

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