Water has a unique property: it reaches its maximum density at 4°C. This means that as water is heated or cooled from 4°C, it expands. Here’s why:

1. If the water is heated above 4°C: It begins to expand, decreasing in density. Since the beaker is already full, this expansion causes water to overflow.

2. If the water is cooled below 4°C: Water also expands when cooled below 4°C due to the formation of an open, hexagonal structure in its molecules (especially as it approaches freezing). This expansion also leads to overflow.

Therefore, the beaker will overflow if the temperature changes from 4°C in either direction: heating above 4°C or cooling below 4°C.

The coefficient of apparent expansion of mercury in a vessel is given by:

\[
\gamma_{\text{apparent}} = \gamma_{\text{mercury}} - \alpha_{\text{vessel}}
\]

For the glass vessel:
\[
153 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{glass}}
\]

For the steel vessel:
\[
144 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{steel}}
\]

Subtracting these two equations, we get:

\[
153 \times 10^{-6} - 144 \times 10^{-6} = \alpha_{\text{steel}} - \alpha_{\text{glass}}
\]

Substitute \(\alpha_{\text{steel}} = 12 \times 10^{-6}/^\circ \text{C}\):

\[
9 \times 10^{-6} = 12 \times 10^{-6} - \alpha_{\text{glass}}
\]

Solving for \(\alpha_{\text{glass}}\):
\[
\alpha_{\text{glass}} = 3 \times 10^{-6}/^\circ \text{C}
\]

Let the length of the other rod be \( L \).

Since the difference in their lengths remains the same at all temperatures, the expansion of each rod must be identical.

For the first rod:
\[
\Delta L_1 = 40 \times \alpha_1 \times \Delta T
\]

For the second rod:
\[
\Delta L_2 = L \times \alpha_2 \times \Delta T
\]

Since \(\Delta L_1 = \Delta L_2\):
\[
40 \times \alpha_1 = L \times \alpha_2
\]

Substitute \(\alpha_1 = 6 \times 10^{-6}/^\circ \text{C}\) and \(\alpha_2 = 4 \times 10^{-6}/^\circ \text{C}\):
\[
40 \times 6 \times 10^{-6} = L \times 4 \times 10^{-6}
\]

Dividing both sides by \(4 \times 10^{-6}\):
\[
L = \frac{40 \times 6}{4} = 60 \, \text{cm}
\]

So, the length of the other rod is \( 60 \, \text{cm} \).

To calculate the heat required to convert 5 g of ice at 0°C to steam at 100°C, we need to consider the following steps:

1. Heat to melt ice (latent heat of fusion):
\[
Q_1 = m \times L_f = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal}
\]

2. Heat to raise temperature of water from 0°C to 100°C:
\[
Q_2 = m \times c \times \Delta T = 5 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - 0) = 5 \times 100 = 500 \, \text{cal}
\]

3. Heat to convert water at 100°C to steam (latent heat of vaporization):
\[
Q_3 = m \times L_v = 5 \, \text{g} \times 540 \, \text{cal/g} = 2700 \, \text{cal}
\]

Total heat required:
\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 400 + 500 + 2700 = 3600 \, \text{cal}
\]

Thus, the total heat required is 3600 calories.

We can solve this problem using the concept of heat exchange. Let \( m_h \) be the mass of hot water at 100°C, and \( m_c \) be the mass of cold water at 10°C. The total mass of water is given as 20 kg, so:

\[
m_h + m_c = 20 \, \text{kg}
\]

We also know that the final temperature of the mixture is 35°C. The heat gained by cold water must equal the heat lost by hot water:

\[
\text{Heat gained by cold water} = \text{Heat lost by hot water}
\]

Let the specific heat of water be \( c = 4200 \, \text{J/kg}^\circ \text{C} \). Using the formula for heat change:

\[
m_c c (35 - 10) = m_h c (100 - 35)
\]

Since the specific heat \( c \) is the same for both, it cancels out:

\[
m_c (35 - 10) = m_h (100 - 35)
\]
\[
m_c \times 25 = m_h \times 65
\]

We also know \( m_h + m_c = 20 \), so \( m_c = 20 - m_h \). Substituting this into the equation:

\[
(20 - m_h) \times 25 = m_h \times 65
\]
\[
500 - 25 m_h = 65 m_h
\]
\[
500 = 90 m_h
\]
\[
m_h = \frac{500}{90} \approx 5.56 \, \text{kg}
\]

Thus, the mass of hot water needed is approximately 5.56 kg.

In the provided graph, the temperature of the material increases with heat input. The flat portions of the graph (AB and CD) represent phase changes where heat is absorbed but the temperature remains constant. Specifically:

- Segment AB corresponds to the latent heat of fusion (solid to liquid).
- Segment CD corresponds to the latent heat of vaporization (liquid to gas).

The length of CD is given as twice the length of AB, meaning the heat input required for vaporization is twice that of fusion.

However, the false conclusion stated in the answer is that "the latent heat of fusion is twice the latent heat of vaporization." This contradicts the graph because CD represents vaporization and is twice the length of AB, implying that the latent heat of vaporization is actually twice the latent heat of fusion, not the other way around.

Hence, the false conclusion is that the latent heat of fusion is larger than the latent heat of vaporization. The correct interpretation from the graph is that the latent heat of vaporization is twice the latent heat of fusion.

Given:
- Mass of water \( m_w = 0.2 \, \text{kg} \)
- Initial temperature of water \( T_w = 80^\circ \text{C} \)
- Final temperature \( T_f = 60^\circ \text{C} \)
- Specific heat of water \( c_w = 4200 \, \text{J/kg}^\circ \text{C} \)
- Mass of solid \( m_s = 0.2 \, \text{kg} \)
- Initial temperature of solid \( T_s = 20^\circ \text{C} \)
- Specific heat of solid \( c_s \) (to be found)

Heat lost by water:
\[
Q_{\text{lost (water)}} = m_w c_w (T_w - T_f) = 0.2 \times 4200 \times (80 - 60) = 16800 \, \text{J}
\]

Heat gained by solid:
\[
Q_{\text{gained (solid)}} = m_s c_s (T_f - T_s) = 0.2 \times c_s \times (60 - 20) = 8 c_s
\]

From the heat exchange equation:
\[
16800 = 8 c_s
\]
\[
c_s = \frac{16800}{8} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Since the specific heat of water is \( 4200 \, \text{J/kg}^\circ \text{C} \), we see that the specific heat of the solid is:

\[
c_s = \frac{4200}{2} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Thus, the specific heat of the solid is indeed half of the specific heat of water.

To find the water equivalent of the calorimeter, we can use the principle of heat exchange:

\[
\text{Heat lost by hotter water} = \text{Heat gained by colder water} + \text{Heat gained by the calorimeter}
\]

Let \( m_1 = 0.2 \, \text{kg} \), \( T_1 = 30^\circ \text{C} \) (colder water), \( m_2 = 0.1 \, \text{kg} \), \( T_2 = 60^\circ \text{C} \) (hotter water), and the final temperature \( T_f = 35^\circ \text{C} \). Let \( W_c \) be the water equivalent of the calorimeter, and \( c = 4200 \, \text{J/kg}^\circ \text{C} \) be the specific heat capacity of water.

Heat lost by hotter water:
\[
Q_{\text{lost}} = m_2 c (T_2 - T_f) = 0.1 \times 4200 \times (60 - 35) = 0.1 \times 4200 \times 25 = 10500 \, \text{J}
\]

Heat gained by colder water:
\[
Q_{\text{gained (water)}} = m_1 c (T_f - T_1) = 0.2 \times 4200 \times (35 - 30) = 0.2 \times 4200 \times 5 = 4200 \, \text{J}
\]

Heat gained by calorimeter:
\[
Q_{\text{gained (calorimeter)}} = W_c \times c \times (T_f - T_1) = W_c \times 4200 \times 5
\]

Using heat balance equation:
\[
Q_{\text{lost}} = Q_{\text{gained (water)}} + Q_{\text{gained (calorimeter)}}
\]
\[
10500 = 4200 + W_c \times 4200 \times 5
\]
\[
10500 - 4200 = 21000 W_c
\]
\[
6300 = 21000 W_c
\]
\[
W_c = \frac{6300}{21000} = 0.3 \, \text{kg}
\]

Since the water equivalent is in terms of mass and \( c = 4200 \, \text{J/kg}^\circ \text{C} \), the water equivalent in \( \text{J/K} \) is:
\[
W_c \times c = 0.3 \times 4200 = 1260 \, \text{J/K}
\]

So, the water equivalent of the calorimeter is 1260 J/K.

To find the ratio of the specific heats of two liquids A and B when they are mixed, we can use the principle of conservation of energy, which states that the heat lost by the hotter liquid equals the heat gained by the colder liquid.

 Given:
- Temperature of liquid A, \( T_A = 32°C \)
- Temperature of liquid B, \( T_B = 24°C \)
- Final temperature of the mixture, \( T_f = 28°C \)

Let \( c_A \) and \( c_B \) be the specific heats of liquids A and B, respectively. Since equal masses of both liquids are mixed, we can denote the mass as \( m \).

 Heat Lost by A:
\[
Q_A = m \cdot c_A \cdot (T_A - T_f) = m \cdot c_A \cdot (32 - 28) = 4m c_A
\]

 Heat Gained by B:
\[
Q_B = m \cdot c_B \cdot (T_f - T_B) = m \cdot c_B \cdot (28 - 24) = 4m c_B
\]

 Setting Heat Lost Equal to Heat Gained:
\[
Q_A = Q_B
\]
\[
4m c_A = 4m c_B
\]

Canceling \( 4m \) from both sides:
\[
c_A = c_B
\]

Conclusion:
The ratio of the specific heats of liquids A and B is:
\[
\frac{c_A}{c_B} = 1
\]

Thus, the ratio of the specific heats is 1:1.

In Part b to c temperature remains constant even when heat is supplied

Similarly in part d to e temperature remains constant

So state is changing in part b to c and in d to e.