(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