(a) Explain specific latent heat
(b)(i) Describe how the specific latent heat of fusion of ice can be determined by the method of mixtures.
(ii) State two precautions to be taken to ensure accurate results.
(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\(^3\) Jg\(^{-1}\), specific latent heat of ice = 3.4 x 10\(^{2}\) Jg\(^{-1}\), specifit heat capacity of water = 4.2 Jg\(^{-1}\) K\(^{-1}\)]
Explanation
(a) Specific latent heat of a substance is the quantity of heat energy required to change a unit mass of a substance from one state to another without change in temperature. The unit specific latent heat is Jkg\(^{-1}\).
(b)(i) Mass of empty calorimeter = M;
Mass of cal. + water = M\(_2\)
Mass of cal. + water + ice = M\(_3\);
Temp. of warm water = \(\theta_1\)
Final temp. when all the is has melted = \(\theta_2\);
Sp. heat capacity of cal = Cc
Sp. heat capacity of water = C\(_w\)
latent heat of ice = L
Mass of water = M\(_2\) - M\(_1\)
Mass of ice = M\(_3\) — M\(_2\)
Heat lost by cal. and water = (m\(_1\)Cc + (M\(_2\) - M\(_1\))C x ( \(\theta_1\) - \(\theta_2\))
Heat gained by ice in melting to water of 0°C = (M\(_3 - M_2\))L
Heat gained by ice water from 0°C = (M\(^3\) - M\(_2\))C\(_ w \theta_2\),
Total heat gained = Heat lost (M\(_3\) - M\(_2\)L + (M\(_3\) - M\(_2\))
C\(_w \theta_2\) = (M\(_1\)Cc + (M\(_2\) - M\(_1\)) C\(_w\) (\(\theta_2 - \theta_2\)
L = \(\frac{M_1C_c + (M_2 - M_1) C_w(\theta_1 - \theta_2) - (M_3 M_2)C_w\theta_2}{(M_3 - M_2)}\)
(ii) Precautions
- Stir gently
- Ensure that ice is dry before adding to water
- Avoid parallax error in reading the balance and thermometer.
(c) If M\(_s\) = mass of steam ; Heat lost by steam at 100°C = M\(_s\)l\(_v\) = M\(_s\) x 2.3 x 10\(^3\)(J)
Heat lost by hot water from 100°C = 0°C ;
= \(M_sC_w \Delta \theta\) = \(M_s\) x 4.2 x 100 (J) Heat gained by ice at 0\(^oC\)
= M\(_{\text{ice}}\) L\(_f\) = 20 x 3.4 x 10\(^2\) = 6800J;
Heat lost by steam + hot water = Heat gained by ice;
M\(_sl_v\) + M\(_s\)C\(_w\)\(\Delta\) = M\(_{\text{ice}}\)L\(_f\);
M\(_s\) (2.3 x 10\(^3\) + 4.2 x 100) = 6800;
M\(_s\) = \(\frac{6800}{2720}\) = 2.5g
Mass of water in container = mass of melted ice + mass of water + mass of condensed steam.
= 20 + 100 + 2.5 = 122.5g