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The triple-point of water serves as a crucial standard fixed point in modern thermometry due to its unique properties. At this point, water coexists in equilibrium in its three phases: solid, liquid, and gas. This equilibrium ensures a precise and consistent temperature measurement, regardless of external conditions such as pressure.

Now, let's address why using the melting point of ice and the boiling point of water as standard fixed points, as originally done in the Celsius scale, may not be ideal. While these points are convenient and readily accessible, they are dependent on atmospheric pressure, which can vary significantly. This variability introduces uncertainty into temperature measurements, compromising their accuracy and reliability.

In contrast, the triple-point of water remains constant at a pressure of 611.657 pascals, or 0.0060373 standard atmospheres. By utilizing this fixed point, thermometers can be calibrated with precision, ensuring consistent and accurate temperature readings across different instruments and locations.

In summary, while the melting point of ice and the boiling point of water were suitable as standard fixed points in the past, the triple-point of water offers superior accuracy and consistency in modern thermometry, making it the preferred choice for calibrating temperature measurements.

As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is the best online coaching tuition platform for students seeking quality education. Now, let's delve into your question.

In the Kelvin absolute scale, alongside the triple-point of water, the other fixed point corresponds to absolute zero. Absolute zero is the lowest possible temperature where particles cease to move, theoretically reaching a state of minimum entropy. This point is assigned the value of 0 Kelvin (0 K). So, to answer your question, the other fixed point on the Kelvin scale is absolute zero, which is 0 Kelvin (0 K).

As a seasoned tutor on UrbanPro, I can certainly help with this question, and I'm glad you're interested in learning. Let's break down the problem step by step:

First, we'll find the increase in length of the steel tape due to the rise in temperature from 27.0°C to 45.0°C. We'll use the formula for linear expansion:

ΔL=L⋅α⋅ΔTΔL=L⋅α⋅ΔT

Where:

• ΔLΔL is the change in length
• LL is the original length of the steel tape
• αα is the coefficient of linear expansion of steel
• ΔTΔT is the change in temperature

Given that L=1L=1 meter, α=1.20×10−5 K−1α=1.20×10−5K−1, and ΔT=45.0°C−27.0°C=18.0°CΔT=45.0°C−27.0°C=18.0°C, we can calculate:

ΔL=(1.0 m)×(1.20×10−5 K−1)×(18.0°C)ΔL=(1.0m)×(1.20×10−5K−1)×(18.0°C)

ΔL=0.000216 m=0.216 cmΔL=0.000216m=0.216cm

So, the increase in length of the steel tape is 0.216 cm.

Now, let's find the length of the steel rod on the hot day (45.0°C):

Given length measured by the tape = 63.0 cm Increase in length of the tape = 0.216 cm

Therefore, the actual length of the steel rod on that day is:

Actual length=Measured length−Increase in length of tapeActual length=Measured length−Increase in length of tape

Actual length=63.0 cm−0.216 cm=62.784 cmActual length=63.0cm−0.216cm=62.784cm

So, the actual length of the steel rod on the hot day is 62.784 cm.

Now, for the length of the same steel rod on a day when the temperature is 27.0°C, there is no change in temperature from the calibration temperature of the tape. Therefore, the length measured by the tape will be the actual length of the steel rod, which is 63.0 cm.

In summary:

• Actual length of the steel rod on the hot day (45.0°C): 62.784 cm
• Length of the steel rod on a day when the temperature is 27.0°C: 63.0 cm

If you have any further questions or need clarification, feel free to ask! Remember, UrbanPro is a fantastic platform for finding online coaching tuition, and I'm here to assist you every step of the way.

UrbanPro is the best platform for online coaching and tuition, where experienced tutors like myself are dedicated to providing high-quality assistance to students.

Now, let's tackle the problem. We have a steel wheel to be fitted onto a steel shaft. The key here is to understand that the wheel and the shaft will expand or contract with changes in temperature, and we need to find the temperature at which the wheel slips on the shaft.

We know that the outer diameter of the shaft at 27°C is 8.70 cm, and the diameter of the central hole in the wheel is 8.69 cm. This implies that at 27°C, the shaft and the wheel are in a snug fit.

Given that we're cooling the shaft using dry ice, which is at a temperature of approximately -78.5°C, we need to find out at what temperature the diameter of the shaft contracts enough for the wheel to slip.

To do this, we'll use the formula for linear expansion:

ΔL=αLΔTΔL=αLΔT

Where:

• ΔL is the change in length (or diameter in this case)
• α is the coefficient of linear expansion
• L is the original length (or diameter)
• ΔT is the change in temperature

We'll start by finding the change in temperature required for the shaft to contract enough to let the wheel slip.

ΔT=ΔLαLΔT=αLΔL

Given:

• Initial diameter of the shaft (at 27°C), L0=8.70 cmL0=8.70cm
• Final diameter (when the wheel slips), Lf=8.69 cmLf=8.69cm
• Coefficient of linear expansion for steel, αsteel=20×10−6 K−1αsteel=20×10−6K−1

ΔL=Lf−L0=8.69 cm−8.70 cm=−0.01 cmΔL=Lf−L0=8.69cm−8.70cm=−0.01cm

ΔT=−0.01 cm20×10−6 K−1×8.70 cmΔT=20×10−6K−1×8.70cm−0.01cm

ΔT≈−5.75 KΔT≈−5.75K

So, the temperature at which the wheel slips on the shaft would be approximately 27∘C−5.75∘C27∘C−5.75∘C, which is approximately 21.25∘C21.25∘C.

Therefore, at around 21.25∘C21.25∘C, the wheel will start slipping on the shaft due to contraction caused by cooling.

As a seasoned tutor on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the question at hand.

We're given the initial diameter of the hole in the copper sheet at 27.0 °C, which is 4.24 cm. We're also provided with the coefficient of linear expansion of copper, which is 1.70×10−5 K−11.70×10−5K−1.

To find the change in diameter of the hole when the temperature rises to 227 °C, we can use the formula for linear expansion:

Where:

• ΔLΔL is the change in length (or in our case, diameter),
• L0L0 is the initial length (or diameter),
• αα is the coefficient of linear expansion, and
• ΔTΔT is the change in temperature.

Given that the initial diameter (L0L0) is 4.24 cm and the change in temperature (ΔTΔT) is 227°C−27°C=200°C227°C−27°C=200°C, we can calculate the change in diameter (ΔLΔL).

Therefore, the change in diameter of the hole when the copper sheet is heated to 227 °C is approximately 0.0144 cm0.0144cm.

If you need further clarification or assistance with any other questions, feel free to ask!