Question

Calculate p.d. at the ends of a power source which, supplies current to a 4 ohm resistance wire for 20 minutes and raises the temperature of 400 g of water through 20°С.

Answer 1

R = 4 Ω; t = 20 min = 1200 s; V = ?

m = 400 g; SHC of water = 4.2 Jg⁻¹°C⁻¹; θ_R = 20°С

Energy produced in wire = \[\frac{V^2 \times t}{R} = \frac{V^2 \times 1200}{4}\text{J}\]

Energy absorbed by water = m c θ_R

= 400 × 4.2 × 20 J

∴ By the law of conservation of energy

\[\frac{V^2 \times 1200}{4} = 400 \times 4.2 \times 20\]

\[\Rightarrow V^2 = \frac{400 \times 4.2 \times 20 \times 4}{1200}\]

\[\Rightarrow V^2 = 112\]

\[\Rightarrow V = \sqrt{112}\]

\[\therefore V = 4\sqrt{7}\text{V} \approx 10.5\text{V}\]