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Using the general gas equation; \(\frac{P_1 \times V_1}{T_1}\) = \(\frac{P_2 \times V_2}{T_2}\) P\(_1\) = P, P\(_2\) = 2P, T\(_1\) = 27ºC → 300K, T\(_2\) = 127ºC → 400k, V\(_2\) = \(\frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}\) V\(_2\) = \(\frac{P \times V_1 \times 400}{300 \times 2P}\) = \(\frac{2V_1}{3}\) \DeltaV = V\(_1\) - \(\frac{2V_1}{3}\) = \(\frac{V_1}{3}\) Therefore, The volume decreases by \(\frac{1}{3}\) of the original volume.

The bursting of water pipes in very cold weather occurs because water expands as it freezes. This expansion increases pressure inside the pipes, which can lead to bursting.

k = \(\frac{Q.d}{A. ΔT .t}\) d = thickness = 3m A = 9m\(^2\) Q = 288kJ = 288000J, t = 7200s, ΔT = 90ºC/m → 90K/m x 3 = 270K k = \(\frac{288000 \times 3}{9 \times 270 \times 7200}\) k = 4.93 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

3\(\alpha\) = \(\gamma\) \(\alpha\) = \(\frac{3.9\times 10^{-6}}{3}\) = 1.3 x 10\(^{-6}\) BUT \(\beta\) = 2\(\alpha\) \(\beta\) = 2 x 1.3 x 10\(^{-6}\) = 2.6 x 10\(^{-6}\)K\(^{-1}\)

Electrical heat generated = quantity of heat required to change water from liquid to vapour I\(^2\) R t = mL I = current = 10A, R = resistance = 2\Omega, t = time = 20s, m = mass of water = 5 X 10\(^{-3}\)kg and L = specific latent heat of vaporization. 10\(^2\) x 2 x 20 = 5 X 10\(^{-3}\)L 4000 = 5 X 10\(^{-3}\)L L = \(\frac{4000}{5\times10^{-3}}\) = 800000 ⇔ 8.0 x 10\(^5\)Jkg\(^{-1}\)

Quantity of heat Q = mass of iron x specific heat capacity of iron x change in temperature. Q = mc\(\delta\)\(\theta\) m = 60g ⇒ 0.060, c = 460Jkg\(^{-1}\)k\(^{-1}\) and \(\delta\)\(\theta\) = 75 - 35 = 40ºC Q = 0.060 x 460 x 40 = 1104J.

Quantity of Heat = heat capacity x change in temperature Q = C\(_p\) x \(\Delta\)\(\theta\) Q = 400 x 20 = 8000J = 8kJ

The quantity of heat required to melt ice = mass of ice x specific latent heat Q = m L = 0.2 x 3.4 x 10\(^5\) = 6.8 x 10\(^4\)J

Relative humidity = \(\frac{\text{S.V.P at dew point}}{\text{S.V.P at air temperature}}\) S.V.P at dew point = 7.3mmHg S.V.P at air temperature = 13.5mmHg R.H = \(\frac{7.3}{13.5}\) x 100 \(\approxeq\) 54%

In a saturated vapour, the rate of evaporation (y) equals the rate of condensation (x). This means that the amount of vapor entering the gaseous phase is equal to the amount of vapour returning to the liquid phase, resulting in a stable concentration of vapour in the air.

Newton's Law of Cooling states that the rate of heat loss of an object is directly proportional to the difference in temperature between the object and its surroundings, provided that this temperature difference is small. Therefore, this law is only valid within a small temperature range.

When thermal energy is increased, the change in state is called melting (solid to liquid) or vaporization (liquid to gas), depending on the initial and final states of the substance.

Gravitational force Fg = \(\frac{Gm_1m_2}{r^2}\) Fg = 10N distance = \(\frac{r}{2}\) F'g = \(\frac{Gm_1m_2}{(\frac{r}{2})^2}\) = 4 (\(\frac{Gm_1m_2}{r^2}\)) = 4Fg = 4 x 10 = 40N

Gravitational force Fg = \(\frac{Gm_1m_2}{r^2}\) = 6.67N \(m_1\) = 10\(^{24}\)kg, \(m_2\) = 10\(^{27}\)kg, G = 6.6 x 10\(^{-11}\)Nm\(^2\)kg\(^{-2}\) and r = ? 6.67 = \(\frac{ 6.6\times 10^{-11}\times10^{24}\times10^{27}}{r^2}\) 6.67r\(^2\) = 6.6 x 10\(^{40}\) r\(^2\) = \(\frac{6.6\times10^{40}}{6.67}\) \(\approx\) 9.89 \times 10\(^{39}\) r = \(\sqrt{ 9.89\times10^{39}}\) \(\approxeq\) 9.94 \times 10\(^{19}\)m → 1.0 x 10\(^{20}\)m

Ebonite rods are negatively charged, while glass rods are positively charged when rubbed. Opposite charges attract each other, leading to an attractive force between the rods.