Two identical beakers `A` and `B` contain equal volumes of two different liquids at `60^(@)C` each and left to cool down. Liquid in A has density of `8xx10^(2)kg//m^(2)` and specific heat of `2000J kg^(-1)K^(-1)` while liquid in `B` has density of `10^(3)kg m^(-3)` and specific heat of `400J kg^(-1)K^(-1)`. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the cooling rates of the two liquids in beakers A and B based on their mass and specific heat capacities. We will use Newton's law of cooling to derive the temperature versus time graph.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Beaker A:
- Density, \( \rho_A = 8 \times 10^2 \, \text{kg/m}^3 \)
- Specific heat, \( c_A = 2000 \, \text{J/(kg K)} \)
- Beaker B:
- Density, \( \rho_B = 10^3 \, \text{kg/m}^3 \)
- Specific heat, \( c_B = 400 \, \text{J/(kg K)} \)
- Both liquids start at \( T_0 = 60^\circ C \).

2. **Calculate the Mass of Liquids:**
- Let the volume of each liquid be \( V \).
- Mass of liquid in beaker A:
\[ m_A = \rho_A \cdot V = (8 \times 10^2) \cdot V = 800V \, \text{kg} \]
- Mass of liquid in beaker B:
\[ m_B = \rho_B \cdot V = (10^3) \cdot V = 1000V \, \text{kg} \]

3. **Calculate the Heat Capacity (m*c) for Each Liquid:**
- Heat capacity for liquid A:
\[ m_A \cdot c_A = (800V) \cdot (2000) = 1.6 \times 10^6 V \, \text{J/K} \]
- Heat capacity for liquid B:
\[ m_B \cdot c_B = (1000V) \cdot (400) = 4.0 \times 10^5 V \, \text{J/K} \]

4. **Determine the Cooling Rate Using Newton's Law of Cooling:**
- According to Newton's law of cooling, the rate of change of temperature is proportional to the temperature difference:
\[ \frac{dT}{dt} \propto (T - T_{\text{ambient}}) \]
- The proportionality constant is inversely related to the heat capacity:
\[ \frac{dT}{dt} = -k \cdot \frac{1}{m \cdot c} \cdot (T - T_{\text{ambient}}) \]
- Therefore, the cooling rate for each liquid can be expressed as:
\[ \frac{dT_A}{dt} = -k_A \cdot (T_A - T_{\text{ambient}}) \quad \text{and} \quad \frac{dT_B}{dt} = -k_B \cdot (T_B - T_{\text{ambient}}) \]
- Where \( k_A \) and \( k_B \) are constants depending on the heat capacity.

5. **Compare the Cooling Rates:**
- Since \( m_A \cdot c_A = 1.6 \times 10^6 V \) and \( m_B \cdot c_B = 4.0 \times 10^5 V \), we see that:
\[ m_A \cdot c_A > m_B \cdot c_B \]
- This implies that the cooling rate for liquid A is slower than that for liquid B:
\[ \left| \frac{dT_A}{dt} \right| < \left| \frac{dT_B}{dt} \right| \]

6. **Conclusion on Temperature vs. Time Graph:**
- Since liquid A cools more slowly than liquid B, the temperature vs. time graph will show that the temperature of liquid A decreases less steeply compared to liquid B.
- Thus, the graph for liquid A will be above that of liquid B for the same time interval.

### Final Answer:
The best schematic representation of the temperature versus time graph is option **B**, where the slope for liquid A is less than that for liquid B.