Two metal rods $1$ and $2$ of same lengths have same temperature difference between their ends. Their thermal conductivities are $ K_1$ and $K_2 $ and cross sectional areas $ A_1 $ and $ A_ 2 $ respectively. If the rate of heat conduction in 1 is four times that in 2, then :

Question

A 200 g sample of water at 80°C is mixed with 100 g of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

Answer 1

To determine which option correctly describes the relationship between the thermal conductivities and cross-sectional areas of the two rods, we'll use the formula for the rate of heat conduction. The rate of heat conduction through a rod is given by Fourier's Law:

Q = \frac{{K \cdot A \cdot \Delta T}}{L}

Where:

Q is the rate of heat transfer (in Watts).
K is the thermal conductivity of the material (in W/m·K).
A is the cross-sectional area of the rod (in m²).
\Delta T is the temperature difference across the ends of the rod (in Kelvin).
L is the length of the rod (in meters).

We know from the problem statement that the length L and the temperature difference \Delta T are the same for both rods. We are also given that the rate of heat conduction in rod 1 is four times that in rod 2, which can be mathematically expressed as:

Q_1 = 4 \cdot Q_2

Substituting the formula for each rod, we get:

\frac{{K_1 \cdot A_1 \cdot \Delta T}}{L} = 4 \cdot \frac{{K_2 \cdot A_2 \cdot \Delta T}}{L}

After canceling out common terms L and \Delta T from both sides, we have:

K_1 \cdot A_1 = 4 \cdot K_2 \cdot A_2

The correct answer is thus K_1A_1 = 4K_2A_2, as it matches the derived equation. This means that the product of the thermal conductivity and cross-sectional area of rod 1 is four times that of rod 2.

Answer 2