### Solution:
### 5. a) Definition
The specific latent heat of vaporization is defined as the amount of heat required to convert 1 kilogram of a substance from its liquid phase to its vapor phase without any change in temperature.
### 5. b) Calculations
#### (i) Heat lost by steam
When steam at \( 100^\circ C \) condenses to form water at \( 100^\circ C \) and then cools down to \( 52^\circ C \), it loses heat in two stages:
1. Condensation of steam:
- Mass of steam: \( 2 \text{ g} = 0.002 \text{ kg} \)
- Latent heat of vaporization (\(L_v\)): \( 2.26 \times 10^6 \text{ J/kg} \)
[tex]\[
Q_{\text{condensation}} = m \cdot L_v = 0.002 \times 2.26 \times 10^6 = 4520 \text{ J}
\][/tex]
2. Cooling of water from \( 100^\circ C \) to \( 52^\circ C \):
- Specific heat capacity of water (\(c\)): \( 4200 \text{ J/kgK} \)
- Temperature change (\(\Delta T\)): \( 100 - 52 = 48^\circ C \)
[tex]\[
Q_{\text{cooling}} = m \cdot c \cdot \Delta T = 0.002 \times 4200 \times 48 = 4032 \text{ J}
\][/tex]
So, the total heat lost by the steam is:
[tex]\[
Q_{\text{total}} = Q_{\text{condensation}} + Q_{\text{cooling}} = 4520 + 4032 = 4923.2 \text{ J}
\][/tex]
#### (ii) Mass of ice used
The heat gained by the ice includes:
1. Melting of ice:
- Latent heat of fusion of ice (\(L_f\)): \( 3.3 \times 10^5 \text{ J/kg} \)
[tex]\[
Q_{\text{melting}} = m_{\text{ice}} \cdot L_f = m_{\text{ice}} \cdot 3.3 \times 10^5
\][/tex]
2. Heating the melted water from \( 0^\circ C \) to \( 52^\circ C \):
- Specific heat capacity of water (\(c\)): \( 4200 \text{ J/kgK} \)
- Temperature change (\(\Delta T\)): \( 52^\circ C \)
[tex]\[
Q_{\text{heating\_water}} = m_{\text{ice}} \cdot c \cdot \Delta T = m_{\text{ice}} \cdot 4200 \cdot 52
\][/tex]
The total heat gained by the ice and water formed from it is:
[tex]\[
Q_{\text{gained}} = Q_{\text{melting}} + Q_{\text{heating\_water}} = m_{\text{ice}} \cdot 3.3 \times 10^5 + m_{\text{ice}} \cdot 4200 \cdot 52
\][/tex]
Setting \( Q_{\text{gained}} \) equal to \( Q_{\text{total}} \):
[tex]\[
4923.2 = m_{\text{ice}} \cdot 3.3 \times 10^5 + m_{\text{ice}} \cdot 4200 \cdot 52
\][/tex]
[tex]\[
4923.2 = m_{\text{ice}} \cdot (3.3 \times 10^5 + 4200 \cdot 52)
\][/tex]
[tex]\[
4923.2 = m_{\text{ice}} \cdot (3.3 \times 10^5 + 218400)
\][/tex]
[tex]\[
4923.2 = m_{\text{ice}} \cdot 548400
\][/tex]
Solving for \( m_{\text{ice}} \):
[tex]\[
m_{\text{ice}} = \frac{4923.2}{548400} \approx 0.008977 \text{ kg}
\][/tex]
Converting to grams:
[tex]\[
m_{\text{ice}} \approx 8.977 \text{ g}
\][/tex]
So, the mass of ice used is approximately 8.977 g.
### 6. Experimental Determination of Specific Heat Capacity
To determine the specific heat capacity of a substance experimentally, you can use a calorimeter and follow these steps:
1. Preparation:
- Take a known mass of the substance whose specific heat capacity is to be measured.
- Measure the initial temperature of the substance.
2. Heating:
- Heat water to a known higher temperature (measure this temperature accurately) and pour it into the calorimeter which already contains the substance.
3. Mixing:
- Stir the mixture thoroughly to ensure proper mixing and measure the final equilibrium temperature of the mixture once no further temperature change is observed.
4. Calculation:
- Apply the principle of conservation of energy, where heat lost by the hot water equals the heat gained by the substance:
[tex]\[
m_w \cdot c_w \cdot (T_{w,\text{initial}} - T_f) = m_s \cdot c_s \cdot (T_f - T_{s,\text{initial}})
\][/tex]
- Here, \(m_w\) and \(m_s\) are the masses of water and substance respectively, \(c_w\) and \(c_s\) are the specific heat capacities of water and the substance respectively, \(T_{w,\text{initial}}\) and \(T_{s,\text{initial}}\) are the initial temperatures of water and the substance, and \(T_f\) is the final equilibrium temperature.
5. Solving:
- Solve for \(c_s\) (specific heat capacity of the substance):
[tex]\[
c_s = \frac{m_w \cdot c_w \cdot (T_{w,\text{initial}} - T_f)}{m_s \cdot (T_f - T_{s,\text{initial}})}
\][/tex]
By accurately measuring the masses, initial and final temperatures, and using the known specific heat capacity of water ([tex]\(4200 \text{ J/kgK}\)[/tex]), you can determine the specific heat capacity of the substance in question.
