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?

Answer 1

The effective resistance of the circuit,

\[R_{eff}  = \left( \frac{6 \times 2}{6 + 2} \right) + 1 = \frac{5}{2}  A\]

Current i through the circuit,

\[i = \frac{V}{R_{eff}} = \frac{6}{5/2} = \frac{12}{5}  A\]

Let i' be the current through the 6 Ω resistor. Then,

i' × 6 = (i − i') × 2

\[\Rightarrow 6i' = \left( \frac{12}{5} \right) \times 2 - 2i'\]

\[ \Rightarrow 8i' = \frac{24}{5}\]

\[ \Rightarrow i' = \frac{24}{(5 \times 8)} = \frac{3}{5}  A\]

\[ \Rightarrow   i - i' = \frac{12}{5} - \frac{3}{5} = \frac{9}{5}  A\]

(a) Heat generated in the 2 Ω resistor,

H = (i - i')²Rt

\[\Rightarrow H = \left( \frac{9}{5} \right) \times \left( \frac{9}{5} \right) \times 2 \times 15 \times 60 = 5832  J\]

The heat capacity of the calorimeter together with water is 2000 J K−1. Thus, 2000 J of heat raise the temp by 1 K.

∴ 5832 J of heat raises the temperature by

\[\frac{5832}{2000}=2.916 K\]

(b) When the 6 Ω resistor gets burnt, the effective resistance of the circuit,

R_eff = 1 + 2 = 3 Ω

Current through the circuit,

i = \[\frac{6}{3}=2A\]

Heat generated in the 2 Ω resistor =  (2)²× 2 × 15 × 60 = 7200 J

2000 J raise the temperature by 1 K.

∴ 7200 J raise the temperature by

\[\frac{7200}{2000}=3.6 K\]