5.
(a) Define specific latent heat of vaporization.

(b) Steam at [tex]100^{\circ}C[/tex] was passed for some time into ice at [tex]0^{\circ}C[/tex]. At the end, the temperature of the water obtained was [tex]52^{\circ}C[/tex] and the increase in mass was [tex]2 \, g[/tex]. Calculate:

(i) Heat lost by steam.

(ii) Mass of ice used.

Take:
- Specific heat capacity of water [tex]= 4200 \, J/(kg \cdot K)[/tex]
- Latent heat of ice [tex]= 3.3 \times 10^5 \, J/kg[/tex]
- Latent heat of vaporization [tex]= 2.26 \times 10^6 \, J/kg[/tex]

6. Describe briefly how you would experimentally determine the specific heat capacity of a substance. (4 marks)



Answer :

### 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.