Molar heat capacity of aluminium is `25 JK^(-1) mol^(-1)`.The heat necessary to raise the temperature of 54 g of aluminium (Atomic mass 27 g `mol^(-1)`) from `30^(@)C` to `50^(@)C` is

Options

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

Explanation

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

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

### Step 2: Calculate the number of moles of aluminum (n)
To find the number of moles (n), we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of aluminum is given as 27 g mol^(-1). Thus,
\[ n = \frac{54 \text{ g}}{27 \text{ g mol}^{-1}} = 2 \text{ moles} \]

### Step 3: Calculate the change in temperature (ΔT)
The change in temperature (ΔT) can be calculated as:
\[ ΔT = T2 - T1 = 50°C - 30°C = 20°C \]
(Note: The change in temperature in Celsius is the same as in Kelvin for temperature differences.)

### Step 4: Calculate the heat (q) required
Using the formula for heat:
\[ q = n \times C \times ΔT \]
Substituting the values we have:
\[ q = 2 \text{ moles} \times 25 \text{ J K}^{-1} \text{ mol}^{-1} \times 20 \text{ K} \]

### Step 5: Perform the multiplication
Calculating this gives:
\[ q = 2 \times 25 \times 20 = 1000 \text{ J} \]

### Step 6: Convert the heat to kilojoules
Since 1 kJ = 1000 J, we can convert the heat:
\[ q = 1000 \text{ J} = 1.0 \text{ kJ} \]

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

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