Question

The surface temperature of the lake is $2^\circ C$. The temperature of the bottom of the lake will be
(A)$2^\circ C$
(B)$3^\circ C$
(C)$4^\circ C$
(D)$5^\circ C$

Answer 1

Hint: The density of an object is defined as mass per unit volume. Similarly, the density of water is the mass/weight of water per unit volume. The density of water depends on temperature. Water shows anomalous behavior which we will discuss here.

Complete step by step solution:
We know, when temperature increases the molecules of an object gain kinetic energy as a result, they move away from each other. Due to this expansion, the density of objects decreases. Similarly, when temperature decreases, the energy is taken from the molecules as a result contraction happens. Due to this density of objects increases.
But in the case of water, anomalous expansion happens when it goes from $4^\circ C$ to $0^\circ C$ instead of contraction and its density decreases.
Now, according to the question, the surface of the lake is at $2^\circ C$. The layer at the bottom will have a density more than the density of the layer at the surface. And we know, water has maximum density at $4^\circ C$. Therefore, the temperature of the bottom layer will be $4^\circ C$.
Hence, option (C) $4^\circ C$is correct.

Note:
This anomalous behavior of water keeps aquatic animals alive even at low temperatures. Beyond $4^\circ C$, ice expands as a result density decreases. Due to this, the temperature of the water at the bottom remains $4^\circ C$.
To study the anomalous behavior of water, hope's apparatus is used.
Due to this anomalous expansion of water, ice floats on water. That is, ice is less dense than water.
Only $\dfrac{1}{{10}}th$ part of the ice floats above the water.