Question Library

(a) The statement: The linear expansivity of a solid is \(1.0 \times 10^{-5}K^{-1}\) means that, A unit length of the solid will expand in length by a fraction \(1.0 \times 10^{-5}\) of the original per kelvin rise in temperature. (Note: \(1.0 \times 10-5 = \frac{1}{100,000}\), a fraction). (b) Experiment to determine the linear expansivity of a steel rod. The apparatus is set up as shown above. The length of the steel rod used is first measured with a meter rule and it is then placed in the jacket with one end fixed. The micrometer screw gauge is turned until the end of the screw just touches the end of the rod, and the gauge reading is noted. It is then turned back enough to make sure that the end of the rod can expand freely. The reading on the thermometer is noted and steam is now passed through the jacket for several minutes. The micrometer is used to note the new reading. More steam is passed again and the reading is taken until a constant reading is obtained. The reading is noted and the temperature is noted from the thermometer. The formula \(\alpha = \frac{l_{f} - l_{i}}{l_{i} (\theta_{f} - \theta_{i})}\) is used to determine the linear expansivity. \(l_{f} - \text{final length} ; l_{i} - \text{initial length} ; \theta_{f} - \text{final temperature} ; \theta_{i} - \text{initial temperature}\) (b)(ii) \(l = 3m ; \theta_{i} = 29^\circC ; \theta_{f} = 41^\circC\) \(\alpha = 1.0 \times 10^{-5} K^{-1}\) Safety gap = change in length. \(\Delta l = \alpha l \Delta \theta\) = \(1 \times 10^{-5} \times 3.0 \times (41 - 29)\) = \(36 \times 10^{-5} = 0.00036m\) (c) Advantages: (1) Used for marking bimetallic strip which is used in thermostat. (2) Removal of tight glass stopper (3) Red-hot rivets in ship building. (4) Expansion of metals is used in bimetallic thermometer. Disadvantages: (1) Expansion of metals or concrete bridges. (2) Cracking of glass cup when hot water is poured into the glass cup. (3) Expansion of balance wheel or wrist-watch. (4) Sagging of overhead wire. (5) Expansion of railway lines.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option E.

Q = mcΔθ = (0.5 x 4200 x 10) + 0.5L L = 334000

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option D.

Heat capacity is the total amount of heat required to change the temperature of an entire object by 1 K (or 1 ^\circC). It depends on both the material and the mass of the object.

20cm .... 100oC 4.5cm .... \(\theta\) 0cm ....0oC \(\frac{4.5 - 0}{20 - 0}\) = \(\frac{0 - 0}{100 - 0}\) \(\theta\) = 22.5oC

Q = mL + mc\(\theta\) = (2 x 333000) + (2 x 2090 x 2) = 674 360J

Heat loss = heat gained mh x ch x \(\theta\)h = mc x cc x \(\theta\)c mh x ch x (\(\theta\) - 20) = mc x cc x (20 - 10) ch = cc ; mc = 3mh (\(\theta\) - 20) = 3( 20 - 10) \(\theta\) = 30 + 20 therefore, \(\theta\) = 50ºC

22,000 = 1.5 x c x (30 - 20) c = 1466Jkg-1 oC-1

Conduction and convection rely on the presence of molecules to transfer heat. In a vacuum, however, there are no molecules available, making these processes impossible.

\(\theta\)p = 2\(\theta\)q mq = 2mp Cp X mp X \(\theta\)p = Cq x mq x \(\theta\)q \(\frac {C_pm_q\theta q}{C_qm_p \theta q}\) = \(\frac {\theta q \times 2mp}{2 \theta qmp}\) \(\frac {C_p}{Cq}\) = \(\frac {2}{2}\)

F = \(\frac{V}{\lambda}\) = \(\frac{3 \times 10^8}{400 \times 10^{-9}}\) = 0.75 x 10\(^{15}\)Hz = 7.5 x 10\(^{14}\) Hz