molar heat capacity of Aluminium is `25 JK^(-1)mol^(-1)`the heat necessary to raise the temperature of 54 gram of aluminium (atomic mass` 27 g mol^(-1)`)from 30°C to 50°C is

Options

A. 1.5kJ
B. 0.5kJ
C. 1.0kJ
D. 2.5kJ

Answer

To solve the problem of calculating the heat necessary to raise the temperature of 54 grams of aluminum from 30°C to 50°C, we will follow these steps:

### Step 1: Identify the given data
- Molar heat capacity of aluminum (C) = 25 J K⁻¹ mol⁻¹
- Mass of aluminum (m) = 54 g
- Atomic mass of aluminum = 27 g mol⁻¹
- Initial temperature (T₁) = 30°C
- Final temperature (T₂) = 50°C

### Step 2: Calculate the number of moles of aluminum (n)
Using the formula:
\[ n = \frac{m}{\text{molar mass}} \]

Substituting the values:
\[ n = \frac{54 \, \text{g}}{27 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 3: Calculate the change in temperature (ΔT)
Using the formula:
\[ \Delta T = T_2 - T_1 \]

Substituting the values:
\[ \Delta T = 50°C - 30°C = 20°C \]
(Note: The change in temperature in Celsius is equivalent to the change in Kelvin.)

### Step 4: Calculate the heat (Q) required
Using the formula:
\[ Q = n \cdot C \cdot \Delta T \]

Substituting the values:
\[ Q = 2 \, \text{mol} \cdot 25 \, \text{J K}^{-1} \text{mol}^{-1} \cdot 20 \, \text{K} \]

Calculating:
\[ Q = 2 \cdot 25 \cdot 20 = 1000 \, \text{J} \]

### Step 5: Convert the heat from Joules to Kilojoules
Since 1 kJ = 1000 J:
\[ Q = \frac{1000 \, \text{J}}{1000} = 1.0 \, \text{kJ} \]

### Final Answer
The heat necessary to raise the temperature of 54 grams of aluminum from 30°C to 50°C is **1.0 kJ**.

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