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The point X is the Absolute zero temperature( the lowest possible temperature). Absolute zero is at -273.15 Celsius or -459.67 Fahrenheit or 0 k.

mass of metal = 50g = 0.05kg, c = 45Jkg\(^{-1}\)k\(^{-1}\), \DeltaT = 80 - 20 = 60ºC Quantity of heat lost(Q) = mass of metal x specific capacity of mass x change in temperature Q = m x c x \DeltaT Q = 0.05 x 45 x 60 = 1350J = 1.35 x 10\(^3\)J

Relative humidity = \(\frac{\text{mass of water vapour in a given volume of air(m)}}{\text{mass of water vapour needed to saturate same volume of air at same temp(M)}}\) x 100% R.H = \(\frac{\text{m}}{\text{M}}\) = \(\frac{0.05}{0.15}\) x 100% Therefore, relative humidity of the air = 33.33%

Given: P = 120W, t = ?, m of water = 1.2kg, c of water = 4200Jkg\(^{-1}\)K\(^{-1}\), \DeltaT = 15ºC Q = pt = m c \DeltaT 120 x t = 1.2 x 4200 x 15 = 75600J t = \(\frac{75600}{120}\) = 630secs → 10.50minutes.

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

Heat lost by the warm water = heat gained by cold water m\(_w\) X c X ( Tw - Tm) = m\(_c\) X c X (Tm - Tc) 200 X ( 70 - 30) = 400 X ( 30 - Tc) 2 x 40 = 120 - 4Tc 4Tc = 120 - 80 = 40 Tc = \(\frac{40}{4}\) = 10ºC

Correct features of a clinical thermometer (that differ from ordinary mercury-in-glass thermometers): - I. A constriction — Yes (the kink just above the bulb prevents mercury from falling back on its own after removal from the body). - III. A short range — Yes (typically 35^\circC to 42^\circC or 43^\circC, suitable only for human body temperature). Thus, only I and III are correct for what makes the clinical thermometer different.

Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\) = \(\frac{80 - 20}{0.02}\) = \(\frac{60}{0.02}\) = 3000Km-1 = 3.0 x 103Km-1

Heat given off by the copper = MCt = 2 x 400 x (100 - 0) Heat absorbed by the ice = ML = M x 3.3 x 10\(^5\) Heat lost = Heat gained ∴ M x 3.3 x 10\(^5\) = 2 x 400 x 100 ∴ M =\(\frac{(2 \times 400 \times 100)}{3.3 \times 10^5}\) = \(\frac{8}{33}\)Kg

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

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

Heat lost = Heat used to melt ice I\(^2\)R t = mL I\(^2\) x 10 x 21 = 10 x 336 I\(^2\) = \(\frac{3360}{210}\) = 16 I = \(\sqrt{16}\) = 4A.

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

8\(\Omega\).....100oC 6\(\Omega\).....\(\theta\)oC 3\(\Omega\).....0oC \(\frac{6 - 3}{8 - 3}\) = \(\frac{\theta - 0}{100 - 0}\) therefore, \(\theta\) = 60oC

\(\frac {V_1}{T_1}\) = \(\frac {V_2}{T_2}\) \(\frac {100}{27 + 273}\) = \(\frac {120}{T + 273}\) T = 87oC