Two identical containers of same emissivity containing liquids `A` & `B` art same temperature of `60^(@)` initially and densitiy `rhoA` and `rhoB` respectively. Where `rhoA lt rhoB`. Which plot best represents the temperature varitation of both with time? Given `(S_(A) = 1000(J)/(kg - K), S_(B) = 2000(J)/(kg- K))`

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the temperature variation of two liquids, A and B, in identical containers over time, given their specific heat capacities and densities.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two liquids A and B, both starting at the same temperature of 60°C.
- The specific heat capacities are given as \( S_A = 1000 \, \text{J/(kg·K)} \) and \( S_B = 2000 \, \text{J/(kg·K)} \).
- The densities are \( \rho_A < \rho_B \), which implies that the mass of liquid A is less than that of liquid B.

2. **Newton's Law of Cooling**:
- According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.
- Mathematically, this can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_0) \]
- Where \( k \) is a constant that depends on the properties of the liquid and the environment.

3. **Finding the Mass of the Liquids**:
- The mass \( m \) of the liquids can be calculated using the formula:
\[ m = \rho \cdot V \]
- Since the containers are identical, we can assume the volume \( V \) is the same for both liquids.
- Therefore, we have:
\[ m_A = \rho_A \cdot V \quad \text{and} \quad m_B = \rho_B \cdot V \]

4. **Analyzing the Rate of Temperature Change**:
- The rate of temperature change can be expressed as:
\[ \frac{dT}{dt} \propto -\frac{1}{m \cdot S} \]
- Since \( S_A < S_B \) and \( m_A < m_B \), we can conclude:
\[ \frac{dT_A}{dt} > \frac{dT_B}{dt} \]
- This means that liquid A will cool down faster than liquid B.

5. **Graphical Representation**:
- Both liquids start at the same temperature (60°C) at \( t = 0 \).
- Since liquid A cools faster, its temperature will decrease more rapidly than that of liquid B.
- Therefore, the plot of temperature versus time will show a steeper decline for liquid A compared to liquid B.

6. **Conclusion**:
- The best representation of the temperature variation of both liquids over time will show liquid A decreasing faster than liquid B, starting from the same initial temperature.

### Final Answer:
The correct plot that represents the temperature variation of both liquids A and B with time is **option B**.