Question

Two identical breakers A and B contain equal volumes of two different liquids at $60^{\circ}C$ each and left to cool down. Liquid in A has density of $8 \times 10^2 \; kg/m^3$ and sp ecific heat o f 2000 J $kg^{-1} \; K^{-1}$ while liquid in B has density o f 103 kg m-3 and sp ecific heat o f 4000 J $kg^{-1} K^{-1}$. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is A

 To solve this problem, we need to understand how the temperature of the liquids in breakers A and B changes over time. The rate at which a liquid cools down depends on its mass, specific heat capacity, and the difference in temperature between the liquid and its surroundings.

Let's analyze each breaker's properties:

• Breaker A:
  • Density of liquid = \(8 \times 10^2 \; \text{kg/m}^3\)
  • Specific heat capacity = 2000 J \(\text{kg}^{-1} \; \text{K}^{-1}\)
• Breaker B:
  • Density of liquid = 103 kg/m³
  • Specific heat capacity = 4000 J \(\text{kg}^{-1} \; \text{K}^{-1}\)

Both liquids are initially at the same temperature of \(60^\circ \text{C}\). Assuming the volumes are equal and the emissivity of both breakers is the same, we can use the properties of heat transfer to understand the cooling process:

The cooling rate depends on the heat capacity (mass × specific heat capacity). The heat capacity is a measure of the energy required to change the temperature of a substance. A higher heat capacity means a slower cooling rate for the same temperature difference.

We calculate the heat capacity for each liquid:

• Heat capacity of liquid in Breaker A:
  • Density of A = \(8 \times 10^2 \; \text{kg/m}^3\)
  • Specific heat of A = 2000 J \(\text{kg}^{-1} \; \text{K}^{-1}\)
  • Heat capacity = Density × Specific Heat
• Heat Capacity of liquid in Breaker B:
  • Density of B = 103 kg/m³
  • Specific heat of B = 4000 J \(\text{kg}^{-1} \; \text{K}^{-1}\)
  • Heat capacity = Density × Specific Heat

Let's simplify this analysis using ratios since the actual volumes and masses are not provided:

Given the higher specific heat of liquid B, even with its lower density, it results in a higher heat capacity compared to liquid A.

Conclusion:

Liquid B, having a higher effective heat capacity, will cool down more slowly compared to liquid A. This implies that the temperature vs. time curve for liquid B will have a gentler slope compared to that of liquid A, meaning that it retains its heat longer.

The correct graph visually represents the slower cooling rate of liquid B compared to liquid A.