Question

The 2.0 Ω resistor shown in the figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J K−1. (a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water? (b) Suppose the 6.0 Ω resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes?
Figure

Answer 1

(a)The rise in the water temperature is 2.916 K;

(b) In the next 15 minutes, the increase in water temperature is 3.6 K.

Explanation:

The circuit’s effective resistance

Reff = 6×26+2+1 = 52 A  

Through the circuit, current i is I = VReff = 125 A

Let the current (i’) pass through 6 Ω resistor. Then, i‘ × 6 = (i − i’) × 2 = 35

A⇒ 95 A

(a) In 2 Ω resistor, the heat generated is H =$\left(i-i^{\prime}\right)^{2} R t$

H = 95×95×2×15×60 = 5832 J

The calorimeter’s heat capacity along with the water is 2000 J $K^{-1}$.

So, the raise in temperature is given by 2.916 K.

(b) When the resistor of 6 Ω is burnt, the circuit’s effective resistance,  

Reff = 1 + 2 = 3 Ω

In the 2 Ω resistor, the heat generated is $2^2$ × 2 × 15 × 60 = 7200 J

By 1K the temperature is raised by 2000 J.

Therefore,

$\frac{7200}{2000}=3.6 \mathrm{K}$