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Chapter 10: Problem 28

On the Celsius scale, the absolute zero of temperature is at (A) \(0^{\circ} \mathrm{C}\) (B) \(-32^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(-273.15^{\circ} \mathrm{C}\)

Short Answer

Expert verified
On the Celsius scale, the absolute zero of temperature is at \(-273.15^{\circ} \mathrm{C}\) (Option D).

Step by step solution

01

Option A: 0掳C

The Celsius scale is defined based on the freezing point of water at 0掳C and the boiling point of water at 100掳C. Absolute zero temperature is the lowest possible temperature, and we know that even water in its solid state is not at its lowest possible temperature. So 0掳C cannot be absolute zero, which rules out option A.
02

Option B: -32掳C

-32掳C is a temperature where certain substances may freeze or reach certain properties, but it's not the lowest possible temperature possible. Absolute zero is the temperature at which all molecular motion theoretically ceases, and -32掳C doesn't meet that criterion. So, option B is not the correct answer.
03

Option C: 100掳C

100掳C is the boiling point of water on the Celsius scale, which is a higher temperature state compared to freezing or other lower temperature phases. Since absolute zero is the lowest possible temperature, 100掳C cannot be the correct answer, ruling out option C.
04

Option D: -273.15掳C

-273.15掳C is the correct value for absolute zero on the Celsius scale. Absolute zero is defined as 0 Kelvin, and the Kelvin scale is related to the Celsius scale through the formula: \(K = 掳C + 273.15 \) To find the Celsius value of absolute zero, we can set K = 0 and solve for 掳C: \(0 = 掳C + 273.15\) \(-273.15 = 掳C\) Since -273.15掳C is the temperature at which all molecular motion would theoretically cease, option D is the correct answer for the absolute zero temperature on the Celsius scale.

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Most popular questions from this chapter

1000 drops of a liquid of surface tension \(\sigma\) and radius \(r\) join together to form a big single drop. The energy released raises the temperature of the drop. If \(\rho\) be the density of the liquid and \(S\) be the specific heat, the rise in temperature of the drop would be \((J=\) Joule's equivalent of heat) (A) \(\frac{\sigma}{J r S \rho}\) (B) \(\frac{10 \sigma}{J r S \rho}\) (C) \(\frac{100 \sigma}{J r S \rho}\) (D) \(\frac{27 \sigma}{10 J r S \rho}\)

As the temperature is increased, the period of pendulum, (A) Increases as its effective length increases even though its centre of mass still remains at the centre of the bob. (B) Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob. (C) Increases as its effective length increases due to shifting to centre of mass below the centre of the bob. (D) Decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.

The radius of a metal sphere at room temperature \(T\) is \(R\) and the coefficient of linear expansion of the metal is \(\alpha\). The sphere heated a little by a temperature \(\Delta T\) so that its new temperature is \(T+\Delta T\). The increase in the volume of the sphere is approximately. (A) \(2 \pi R \alpha \Delta T\) (B) \(\pi R^{2} \alpha \Delta T\) (C) \(4 \pi R^{3} \alpha \Delta T / 3\) (D) \(4 \pi R^{3} \alpha \Delta T\)

Boyle's law is applicable for an (A) adiabatic process. (B) isothermal process. (C) isobaric process. (D) isochoric process.

Two litres of water at initial temperature of \(27^{\circ} \mathrm{C}\) is heated by a heater of power \(1 \mathrm{~kW}\) in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of \(160 \mathrm{~J} / \mathrm{s}\). The time in which the temperature will rise from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\) is (specific heat of water \(=4.2 \mathrm{~kJ} / \mathrm{kg})\) (A) \(5 \min 20 \mathrm{~s}\) (B) \(8 \min 20 \mathrm{~s}\) (C) \(10 \min 40 \mathrm{~s}\) (D) \(12 \min 50 \mathrm{~s}\)
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