Problem 46 - Arihant AIEEE Physics

Question

A drilling machine of \(10 \mathrm{~kW}\) power is used to drill a bore in a small aluminium block of mass \(8 \mathrm{~kg}\). If \(50 \%\) of power is used up in heating the machine itself or lost to the surroundings then how much is the rise in temperature of the block in \(2.5\) minutes? (Given: specific heat of aluminium \(=0.91 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\) ) (a) \(103^{\circ} \mathrm{C}\) (b) \(130^{\circ} \mathrm{C}\) (c) \(105^{\circ} \mathrm{C}\) (d) \(30^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The rise in temperature of the block is approximately \(103^{\circ} \mathrm{C}\).

Step 1: Convert Mass to Grams

The mass of the aluminum block is given as \(8\, \mathrm{kg}\). Convert this to grams because the specific heat is given in \(\mathrm{J/g^\circ C}\). \[8\, \mathrm{kg} = 8000\, \mathrm{g}\]

Step 2: Calculate Usable Power for Heating

The total power of the drilling machine is \(10\, \mathrm{kW}\). Since only \(50\%\) is available for heating the aluminum, the usable power is:\[5\, \mathrm{kW} = 5000\, \mathrm{W}\]

Step 3: Calculate Total Energy Transferred

To find the total energy transferred to the aluminum block, use the formula for power:\[\text{Power} = \frac{\text{Energy}}{\text{Time}}\]Rearranging gives:\[\text{Energy} = \text{Power} \times \text{Time}\]The time given is \(2.5\, \mathrm{minutes}\), which is \(150\, \mathrm{seconds}\) (since there are 60 seconds in a minute). Thus, the energy transferred is:\[\text{Energy} = 5000\, \mathrm{W} \times 150\, \mathrm{s} = 750,000\, \mathrm{J}\]

Step 4: Use the Heat Transfer Equation

Use the heat transfer equation:\[Q = mc\Delta T\]Where:- \(Q\) is the total energy transferred, \(750,000\, \mathrm{J}\).- \(m\) is the mass, which is \(8000\, \mathrm{g}\).- \(c\) is the specific heat of aluminum, \(0.91\, \mathrm{J/g^\circ C}\).- \(\Delta T\) is the change in temperature, which we need to find.Rearrange the equation for \(\Delta T:\)\[\Delta T = \frac{Q}{mc} = \frac{750,000}{8000 \times 0.91}\]

Step 5: Calculate the Temperature Change

Plug in the values and calculate \(\Delta T\):\[\Delta T = \frac{750,000}{7280} \approx 103.02^{\circ} \mathrm{C}\].Therefore, the rise in temperature of the block is approximately \(103^{\circ} \mathrm{C}\).

Key Concepts

Power and Energy Conversion

Understanding how power and energy conversion works is essential when dealing with machines like drills. Power is the rate at which energy is used or transferred. In our exercise, the drilling machine has a power rating of 10 kW, meaning it can convert or transfer 10,000 joules of energy per second.

However, not all of this power is used effectively. In real-world scenarios, a portion of energy is often lost to the surroundings or other components within the system. In our case, 50% of the power is lost which leaves us with only 5 kW for heating the aluminium block.

• Total Power: 10 kW (or 10,000 watts)
• Usable Power: 5 kW (or 5,000 watts)

To find out how much energy is transferred, multiply the usable power by the time the machine operates. In the exercise, the drill runs for 2.5 minutes, which is 150 seconds. So, the energy transferred to the block is 750,000 joules. This value is crucial for calculating subsequent changes, such as temperature change.

Temperature Change

The temperature change in an object when it absorbs heat can be calculated using the heat transfer equation:

• \( Q = mc\Delta T \)

Here, \( Q \) is the total heat energy transferred, \( m \) is the mass of the object, \( c \) is its specific heat capacity, and \( \Delta T \) is the change in temperature we wish to find.

In this scenario, after determining the total energy transferred (750,000 J), the mass of the aluminium is converted into grams (8000 g), and the specific heat capacity of aluminium is already provided (0.91 J/g°C). Thus, the formula rearranges to:

• \( \Delta T = \frac{Q}{mc} \)

Plugging in the values: \( \Delta T = \frac{750,000}{8000 \times 0.91} \approxeq 103.02^{\circ} \text{C} \). This temperature change signifies how much hotter the block will get during the drilling.

Aluminium Properties

Aluminium is chosen for many applications due to its unique properties. It's lightweight, yet strong, and is an excellent conductor of heat. Its specific heat capacity is relatively high at 0.91 J/g°C, which indicates it can absorb a substantial amount of heat before its temperature rises significantly.

This property plays an important role in the temperature rise calculation using the drill. The specific heat capacity (c) is crucial to determine how much energy is required to change the temperature of a mass of aluminium by one degree Celsius.

If the aluminium block were made of a material with a lower specific heat capacity, it would require less energy for the same temperature increase. Conversely, materials with higher specific heat capacities need more energy. The operation with such specifics gives us important insights into how different materials behave in thermal energy scenarios, helping us choose suitable materials for heat management.