A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. Find the rise in temperature of the block in `2.5` minutes, assuming `50%` power is used up in heating the machine itself or lost to the surropundings.
(Specific heat of aluminium `= 0.91 J g^(-1).^(@)C^(-1)`)

Options

A. `100 .^(@)C`
B. `103 .^(@)C`
C. `150 .^(@)C`
D. `155 .^(@)C`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Write down the given data
- Power of the drilling machine, \( P = 10 \, \text{kW} = 10 \times 10^3 \, \text{W} = 10^4 \, \text{W} \)
- Mass of the aluminum block, \( m = 8 \, \text{kg} \)
- Time of operation, \( t = 2.5 \, \text{minutes} = 2.5 \times 60 \, \text{s} = 150 \, \text{s} \)
- Specific heat of aluminum, \( c = 0.91 \, \text{J/g}^\circ \text{C} = 910 \, \text{J/kg}^\circ \text{C} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \))

### Step 2: Calculate the total energy produced by the machine
The total energy produced by the machine in the given time is calculated using the formula:
\[ \text{Total Energy} = P \times t \]
Substituting the values:
\[ \text{Total Energy} = 10^4 \, \text{W} \times 150 \, \text{s} = 1.5 \times 10^6 \, \text{J} \]

### Step 3: Calculate the energy available for heating the aluminum block
Since 50% of the energy is lost, the energy available for heating the aluminum block is:
\[ \Delta Q = \frac{1}{2} \times \text{Total Energy} = \frac{1}{2} \times 1.5 \times 10^6 \, \text{J} = 0.75 \times 10^6 \, \text{J} = 7.5 \times 10^5 \, \text{J} \]

### Step 4: Use the formula for heat transfer to find the rise in temperature
The formula relating heat transfer to mass, specific heat, and temperature change is:
\[ \Delta Q = m \cdot c \cdot \Delta T \]
Rearranging this to find the change in temperature (\( \Delta T \)):
\[ \Delta T = \frac{\Delta Q}{m \cdot c} \]
Substituting the values:
\[ \Delta T = \frac{7.5 \times 10^5 \, \text{J}}{8 \, \text{kg} \times 910 \, \text{J/kg}^\circ \text{C}} \]
Calculating the denominator:
\[ 8 \times 910 = 7280 \, \text{J/}^\circ \text{C} \]
Now substituting this back into the equation:
\[ \Delta T = \frac{7.5 \times 10^5}{7280} \approx 102.9 \, \text{°C} \]

### Step 5: Final answer
The rise in temperature of the aluminum block is approximately:
\[ \Delta T \approx 103 \, \text{°C} \]