Solution:

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.

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Solution:

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.

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Solution:

1. Heat gained by ice to reach 0°C:
\[
Q_{\text{ice heating}} = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20°C = 100 \, \text{cal}
\]

2. Heat gained by ice to melt:
\[
Q_{\text{ice melting}} = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal}
\]

Total heat gained by ice:
\[
Q_{\text{total ice}} = 100 \, \text{cal} + 800 \, \text{cal} = 900 \, \text{cal}
\]

3. Heat lost by water:
\[
Q_{\text{water cooling}} = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (50 - T) = 500 - 10T \, \text{cal}
\]

4. Set heat gained equal to heat lost:
\[
500 - 10T = 900 \Rightarrow -10T = 400 \Rightarrow T = -40°C
\]

Since this temperature is below 0°C, all the ice will not melt.

5. Equilibrium Calculation:
Let \( x \) be the mass of ice that melts:
\[
500 = 70x + 100 \Rightarrow 400 = 70x \Rightarrow x \approx 5.71 \, \text{g}
\]

Result:
The final temperature of the system will be **0°C**.

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Solution:

Given:

- Diameter ratio = 1:2
- Density ratio = 2:1
- Specific heat ratio = 1:3

Step 1: Mass ratio
Mass is proportional to \( \text{density} \times \text{volume} \).
Volume is proportional to the cube of the diameter, so the volume ratio is \( (1:2)^3 = 1:8 \).

Thus, mass ratio = \( \text{density} \times \text{volume} = \frac{2 \times 1}{1 \times 8} = 1:4 \).

Step 2: Thermal capacity ratio
Thermal capacity = mass × specific heat.

So, thermal capacity ratio = \( \frac{1 \times 1}{4 \times 3} = 1:12 \).

Final Answer:
The ratio of the thermal capacities is 1:12.

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Solution:

To find the resulting temperature when two liquids A and B are mixed, we can use the principle of conservation of heat energy: the heat lost by the hotter liquid (A) equals the heat gained by the cooler liquid (B).

Given:
- Temperature of liquid A = 75°C
- Temperature of liquid B = 15°C
- Masses of A and B are in the ratio 2:3, so let the masses of A and B be \( 2m \) and \( 3m \), respectively.
- Specific heats of A and B are in the ratio 3:4, so let the specific heats of A and B be \( 3c \) and \( 4c \), respectively.

Let the final temperature of the mixture be \( T \).

Step 1: Heat lost by liquid A (hotter liquid):
\[
Q_A = \text{mass of A} \times \text{specific heat of A} \times (\text{initial temp of A} - T)
\]
\[
Q_A = 2m \times 3c \times (75 - T) = 6mc(75 - T)
\]

Step 2: Heat gained by liquid B (cooler liquid):
\[
Q_B = \text{mass of B} \times \text{specific heat of B} \times (T - \text{initial temp of B})
\]
\[
Q_B = 3m \times 4c \times (T - 15) = 12mc(T - 15)
\]

Step 3: Apply the conservation of heat:
Heat lost by A = Heat gained by B
\[
6mc(75 - T) = 12mc(T - 15)
\]

 Step 4: Simplify and solve for \( T \):
Cancel out \( mc \) from both sides:
\[
6(75 - T) = 12(T - 15)
\]
Expand both sides:
\[
450 - 6T = 12T - 180
\]
Combine like terms:
\[
450 + 180 = 12T + 6T
\]
\[
630 = 18T
\]
Solve for \( T \):
\[
T = \frac{630}{18} = 35 \, ^\circ\text{C}
\]

Final Answer:
The resulting temperature is 35°C.

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Solution:

Here’s a short solution:

Given

- Mass of water = 54 gm
- Initial temperature of water = 30°C, Final temperature = 90°C
- Latent heat of steam = 530 cal/gm
- Specific heat of water = 1 cal/gm°C

Step 1: Heat gained by water to reach 90°C:

\[
Q_{\text{gained}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T = 54 \times 1 \times (90 - 30) = 3240 \, \text{cal}
\]

Step 2: Heat lost by steam:

\[
Q_{\text{lost}} = m_s \times (530 + 10) = 540 m_s
\]

Step 3: Equating heat gained and heat lost:

\[
3240 = 540 m_s \quad \Rightarrow \quad m_s = \frac{3240}{540} = 6 \, \text{gm}
\]

Step 4: Total mass of the mixture:

\[
m_{\text{mixture}} = 54 \, \text{gm} + 6 \, \text{gm} = 60 \, \text{gm}
\]

Final Answer:
The mass of the mixture is 60 grams.

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Solution:

If the portion AB is slant, it indicates that both pressure (P) and volume (V) are changing. In thermodynamic terms, this suggests that AB could represent a compression or expansion, process where both pressure decreases and volume decreases.

Given that you mentioned that the correct answer is "the liquid state of matter," it is likely that AB represents the compression of a liquid.

In many thermodynamic diagrams, when a liquid is compressed or expands slightly, both its pressure and volume can change, though not as dramatically as in gases. This phase corresponds to the liquid region of a substance's phase diagram, where both pressure and volume are decreasing, which is why AB represents the liquid state.

To summarize:
- AB is slant: Both pressure and volume are decreasing.
- Liquid state: Liquids are not perfectly incompressible; hence, both pressure and volume can change, but the changes in volume are relatively small compared to gases.

This slant line indicates the behaviour of a liquid as it is compressed, leading to the conclusion that AB represents the liquid state of matter.

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Solution:

1. Heat released by \(x\) gm steam at \(100^\circ C\):
\[
Q_{\text{steam}} = x \times 540
\]

2. Heat required to convert \(y\) gm ice at \(0^\circ C\) to water at \(100^\circ C\):
\[
Q_{\text{ice}} = y \times 80 + y \times 100 = y \times 180
\]

3. Equating:
\[
x \times 540 = y \times 180
\]

\[
\frac{y}{x} = 3
\]

So, \( y : x = 3 : 1 \).

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