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Step-by-Step Solution
Step 1: Identify the Physical Quantities
Two liquids (A and B) are at the same initial temperature (60°C) and have the same volume. They differ in density ($\rho$) and specific heat capacity ($s$). We assume they cool by losing heat at a rate proportional to the temperature difference $(T - T_0)$ with the same emissivity for both beakers.
Step 2: Relate Mass to Density and Volume
The mass $m$ of each liquid is given by
$$
m = V \cdot \rho
$$
where $V$ is the volume and $\rho$ is the density.
Step 3: Express the Rate of Cooling
When a liquid cools, the rate of decrease of temperature is inversely proportional to its total heat capacity ($m \cdot s$). Symbolically:
$$
\frac{dQ}{dt}
\propto \frac{1}{m\,s}
=
\frac{1}{V \,\rho \,s}.
$$
Since $V$ is the same for both liquids, the rate of cooling depends on $1/(\rho\,s)$. A larger value of $\rho\,s$ implies a smaller cooling rate.
Step 4: Compare the Product $\rho\,s$ for Each Liquid
Liquid A:
$$
\rho_A = 8 \times 10^2 \,\text{kg\,m}^{-3},
\quad
s_A = 2000 \,\text{J\,kg}^{-1}\,\text{K}^{-1}.
$$
$$
\rho_A s_A = (8 \times 10^2) \times 2000 = 1.6 \times 10^6.
$$
Liquid B:
$$
\rho_B = 10^3 \,\text{kg\,m}^{-3},
\quad
s_B = 4000 \,\text{J\,kg}^{-1}\,\text{K}^{-1}.
$$
$$
\rho_B s_B = (10^3) \times 4000 = 4 \times 10^6.
$$
Hence,
$$
\rho_A s_A < \rho_B s_B.
$$
Therefore,
$$
\frac{1}{\rho_A s_A} > \frac{1}{\rho_B s_B}.
$$
Step 5: Conclude Which Liquid Cools Faster
Since the cooling rate is proportional to $1/(\rho\,s)$, Liquid A (with the smaller product $\rho_A s_A$) has a larger cooling rate than Liquid B. Consequently, Liquid A’s temperature falls faster, leading to a steeper slope in the temperature vs. time graph.
Final Answer about the Graph
Liquid B’s graph remains above Liquid A’s graph as time progresses because Liquid B cools more slowly. The correct temperature vs. time graph shows the curve for B above that for A, with A having the steeper slope.
