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^3` and specific heat of `2000 J kg ^(-1)K^(-1)` while liquid in B has density of `10^(3)kgm^(-3)` and specific heat of `4000J 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.

Explanation

To solve the problem, we need to analyze the cooling rates of the two liquids in beakers A and B based on their specific heat capacities and masses. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Mass of the Liquids
The mass of the liquid in each beaker can be calculated using the formula:
\[ \text{Mass} (m) = \text{Density} (\rho) \times \text{Volume} (V) \]

- For beaker A:
\[ m_A = \rho_A \times V = (800 \, \text{kg/m}^3) \times V \]

- For beaker B:
\[ m_B = \rho_B \times V = (1000 \, \text{kg/m}^3) \times V \]

### Step 2: Identify the Specific Heat Capacities
The specific heat capacities of the liquids are given as:
- For liquid A: \( c_A = 2000 \, \text{J/kg/K} \)
- For liquid B: \( c_B = 4000 \, \text{J/kg/K} \)

### Step 3: Calculate the Heat Capacity (C) of Each Liquid
The heat capacity \( C \) is calculated as:
\[ C = m \times c \]

- For beaker A:
\[ C_A = m_A \times c_A = (800V) \times (2000) = 1600000V \, \text{J/K} \]

- For beaker B:
\[ C_B = m_B \times c_B = (1000V) \times (4000) = 4000000V \, \text{J/K} \]

### Step 4: Compare the Heat Capacities
From the calculations:
- \( C_A = 1600000V \)
- \( C_B = 4000000V \)

Since \( C_B > C_A \), it indicates that beaker B has a higher heat capacity than beaker A.

### Step 5: Apply Newton's Law of Cooling
According to Newton's law of cooling, the rate of change of temperature is given by:
\[ \frac{d\theta}{dt} = -k \frac{(T - T_0)}{C} \]

Where:
- \( T \) is the temperature of the liquid,
- \( T_0 \) is the surrounding temperature,
- \( k \) is a constant.

Since \( C_B > C_A \), the rate of temperature change \( \frac{d\theta}{dt} \) will be smaller for beaker B compared to beaker A.

### Step 6: Analyze the Temperature vs. Time Graph
Since the cooling rate of beaker B is slower than that of beaker A, the temperature of liquid A will decrease faster than that of liquid B. Therefore, the temperature vs. time graph will show:
- Beaker A will have a steeper slope (cooling faster),
- Beaker B will have a gentler slope (cooling slower).

### Conclusion
The best schematic representation of the temperature versus time graph will show the temperature of liquid A decreasing faster than that of liquid B.