91Ó°ÊÓ

A motor supplies \(0.4\) hp to stir \(5 \mathrm{~kg}\) of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water \(6{ }^{\circ} \mathrm{C}\) ? The heat required to heat the water is $$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$ This is actually supplied by friction work, so $$\text { Friction work done }=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}$$ and this equals the work done by the motor. But Work done by motor in time \(t=(\) Power \()(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)\) Equating this to our previous value for the work done yields $$ t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min} $$

Step by step solution

01

Calculate Heat Required

First, calculate the heat required to raise the temperature of 5 kg (5000 g) of water by 6°C using the specific heat capacity. The formula is \( \Delta Q = m \cdot c \cdot \Delta T \), where \( c = 1 \text{ cal/g}\cdot^\circ\text{C} \). Therefore, \( \Delta Q = 5000 \text{ g} \times 1 \text{ cal/g}\cdot^\circ\text{C} \times 6^\circ\text{C} = 30\text{ kcal} \).

02

Convert Heat to Joules

Convert the heat from calories to joules. Since 1 calorie = 4.184 Joules, the energy in joules is \( 30 \text{kcal} \times 4.184 \text{J/cal} = 126 \text{kJ} \).

03

Determine Motor Work

The work done by the motor is equivalent to the heat required for the water heating, which is \( 126\text{kJ} \).

04

Calculate Power Output

Convert the power supply of the motor into watts. The motor supplies \( 0.4 \text{ hp} \), and \( 1 \text{ hp} = 746 \text{ W} \). Thus, power in watts is \( 0.4 \text{ hp} \times 746 \text{ W/hp} = 298.4 \text{ W} \).

05

Calculate Time Required

Using the formula \( \text{Work} = \text{Power} \times \text{Time} \), solve for time \( t \). Using \( \text{Work} = 126000 \text{ J} \) and \( \text{Power} = 298.4 \text{ W} \), we have \( t = \frac{126000 \text{ J}}{298.4 \text{ W}} \approx 422 \text{ s} \).

06

Convert Time to Minutes

Convert the time required from seconds to minutes: \( 422 \text{ s} \approx 7 \text{ min} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one body or substance to another. There are different mechanisms through which heat can be transferred, such as conduction, convection, and radiation. In this exercise, the heat transfer occurs because of friction, as mechanical energy from the motor is converted into thermal energy, increasing the water's temperature.
The process relies on energy conservation principles, where the mechanical work done by the motor becomes the thermal energy absorbed by the water. Understanding heat transfer is essential for recognizing how energy changes form and moves within different systems.

Specific Heat Capacity

Specific heat capacity is a property that defines how much energy is needed to change the temperature of one gram of a substance by one degree Celsius. Water has a high specific heat capacity, which means it requires a significant amount of energy to change its temperature.
In our exercise, the specific heat capacity of water is used to calculate the energy required to raise the temperature by 6°C. The calculation involves multiplying the mass of the water, the change in temperature, and the specific heat capacity of water (1 cal/g·°C). This multiplication gives the total energy needed, helping to understand how the substance's properties affect heat transfer.

Unit Conversions

Unit conversions are crucial in scientific calculations to maintain consistency and accuracy. In the given exercise, different units such as calories, joules, and horsepower must be converted appropriately. This ensures that all parts of the calculation are compatible.
The specific heat calculation initially yields energy in kilocalories, but since work and power are expressed in joules and watts, it's essential to convert kilocalories to joules using 1 cal = 4.184 J. Similarly, horsepower is converted to watts to determine the motor's power output. Proper unit conversion is critical for obtaining accurate results in physics calculations.

Mechanical Work

Mechanical work is done when a force is applied to an object, causing it to move. In the context of this exercise, the motor performs work by stirring the water, increasing its temperature due to friction. The work done is equal to the energy transferred to the water.
When dealing with mechanical work, it is important to relate it to power and time using the formula: \( \text{Work} = \text{Power} \times \text{Time} \). Completing the calculation with correct power and energy values lets us find the time it takes for the motor to heat the water. Understanding mechanical work and its relation to energy and power is pivotal in solving similar practical physics problems.