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Heat lost = heat gained 200 x c x (90 - \(\theta\))= 100\(\times\) c \(\times\) (\(\theta - 30\)) Dividing through by 100c, we have 2(\(90 - \theta\)) = (\(\theta - 30\)) simplifying, we have \(\theta\) = 70^\circ

Gap ΔL = αL1Δθ = 1.0 x 10-5 x 3 x (40 - 28) = 3.6 x 10-2

Using Charles law: V1/T1 = V2/T2; (283 * 303) / 283 V2= 303

I.Vt = mcθ; C = I.Vt/mθ = 75 x 4 x 60/1.5 x 20 = 6.0 X102J kg-10C-1

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 D.

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

\(\alpha\) = \(\frac{extension}{L\Delta\theta}\) 0.000012 = \(\frac{e}{600 \times (35 - 10}\) = 0.18m = (0.18 x 100) cm = 18.0 cm

Q = mc\(\theta\) = 0.05 x 390 x (70 - 20) = 975J

The e.m.f. is primarily influenced by temperature, the nature of the plates, and the nature of the electrolyte, but not by the size of the cell itself.

100oC .... 5.5 \(\theta\)oC .... 5.3 0oC .... 5.0 \(\frac{\theta - 0}{100 - 0}\) = \(\frac{5.5 - 5 .0}{5.5 - 5.0}\) \(\theta\) = 40oC

100oC ... 90oS \(\theta\)oC ... 60oS 0oC ... 30oS \(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\) \(\theta\) = 75oC

\(\alpha\) = \(\frac{L_2 - L_1}{L_1 \times \Delta \theta}\) 18 x 10-6 = \(\frac{L_2 - 2}{2\times 100}\) L = 2.0036m

To find the difference in the amount of heat given out by 4 kg of steam and 4 kg of water when both are cooled from 100^\circC to 80^\circC, we need to calculate the heat released by each. Heat Given Out by Water The heat given out by water as it cools can be calculated using the formula: \(Q = mc\Delta T\) Where: - m = 4 kg (mass of water) - c = 4200 (specific heat capacity of water) - \( \Delta\) T = 100 - 80 = 20K (change in temperature) Calculating the heat given out by water: \(Q_{\text{water}} = 4 \times 4200 \times 20 = 336,000 \, \text{J}\) Heat Given Out by Steam For steam, we need to consider both the cooling of the steam to water and the cooling of the water from 100^\circC to 80^\circC. The total heat given out by the steam is the sum of the heat released during condensation and the heat released during cooling. Heat released during condensation: \(Q_{\text{condensation}} = mL\) Where: - \( L = 2,260,000 \, \text{J/kg} \) (specific latent heat of steam) Calculating the heat released during condensation: \(Q_{\text{condensation}} = 4 \, \text{kg} \times 2,260,000 \, \text{J/kg} = 9,040,000 \, \text{J}\) Heat released by water as it cools from 100^\circC to 80^\circC: This is the same calculation we did for water above: \(Q_{\text{cooling}} = 4 \times 4200 \times 20 = 336,000 \, \text{J}\) Total heat given out by steam: \(Q_{\text{steam}} = Q_{\text{condensation}} + Q_{\text{cooling}} = 9,040,000 \, \text{J} + 336,000 \, \text{J} = 9,376,000 \, \text{J}\) Difference in Heat Given Out Now, we find the difference in heat given out by steam and water: \(\text{Difference} = Q_{\text{steam}} - Q_{\text{water}} = 9,376,000 \, \text{J} - 336,000 \, \text{J} = 9,040,000 \, \text{J}\) The difference in the amount of heat given out by 4 kg of steam and 4 kg of water when both are cooled from 100^\circC to 80^\circC is 9,040,000 J.

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