A `10 kW` drilling machine is used to drill a bore in a small aluminium block of mass `8.0 kg`. How much is the rise in temperature of the block in `2.5` minutes, assuming `50%` of power is used up in heating the machine itself or lost to the surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`.

Options

A. `83^(@)C`
B. `103^(@)C`
C. `150^(@)C`
D. `123^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the total energy delivered by the drilling machine.
The power of the drilling machine is given as \(10 \, \text{kW}\), which is equivalent to \(10,000 \, \text{W}\) or \(10^4 \, \text{J/s}\). The time for which the machine is used is \(2.5 \, \text{minutes}\).

Convert time from minutes to seconds:
\[ 2.5 \, \text{minutes} = 2.5 \times 60 \, \text{seconds} = 150 \, \text{seconds} \]

Now, calculate the total energy delivered:
\[ \text{Total Energy} = \text{Power} \times \text{Time} = 10^4 \, \text{J/s} \times 150 \, \text{s} = 1.5 \times 10^6 \, \text{J} \]

### Step 2: Calculate the energy used for heating the aluminum block.
Since \(50\%\) of the power is lost to the surroundings or used by the machine itself, the energy used for heating the aluminum block is:
\[ \text{Energy for heating} = 0.5 \times \text{Total Energy} = 0.5 \times 1.5 \times 10^6 \, \text{J} = 7.5 \times 10^5 \, \text{J} \]

### Step 3: Use the specific heat formula to find the rise in temperature.
The specific heat of aluminum is given as \(0.91 \, \text{J/g°C}\). To use this in our calculations, we need to convert the mass of the aluminum block from kilograms to grams:
\[ \text{Mass} = 8.0 \, \text{kg} = 8000 \, \text{g} \]

The formula for heat energy is:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \(Q\) = heat energy (in Joules)
- \(m\) = mass (in grams)
- \(c\) = specific heat capacity (in J/g°C)
- \(\Delta T\) = change in temperature (in °C)

Rearranging the formula to solve for \(\Delta T\):
\[ \Delta T = \frac{Q}{m \cdot c} \]

Substituting the known values:
\[ \Delta T = \frac{7.5 \times 10^5 \, \text{J}}{8000 \, \text{g} \times 0.91 \, \text{J/g°C}} \]

Calculating the denominator:
\[ 8000 \, \text{g} \times 0.91 \, \text{J/g°C} = 7280 \, \text{J/°C} \]

Now substituting back into the equation for \(\Delta T\):
\[ \Delta T = \frac{7.5 \times 10^5 \, \text{J}}{7280 \, \text{J/°C}} \approx 102.9 \, °C \]

### Final Answer:
The rise in temperature of the aluminum block is approximately \(102.9 \, °C\).

---